# Planar Brownian Motion and Complex Analysis

Greg Markowsky

###### Contents

  1. 0 Preface
  2. 1 Preliminaries
    1. 1.1 Basic notions from probability theory
    2. 1.2 Brownian motion
    3. 1.3 Complex numbers
    4. 1.4 Exercises
  3. 2 Dynkin’s formula and Lévy’s Theorem
    1. 2.1 Dynkin’s formula
    2. 2.2 Analytic invariance of Brownian motion
    3. 2.3 Möbius transformations
    4. 2.4 Poisson’s integral formula and power series
    5. 2.5 Exercises
  4. 3 Examples and basic properties of harmonic and analytic functions
    1. 3.1 General properties of harmonic functions
    2. 3.2 The Dirichlet Problem
    3. 3.3 The Identity Theorem, Open Mapping Theorem, and Maximum Principle
    4. 3.4 The exponential and trigonometric functions
    5. 3.5 Entire and almost entire functions
    6. 3.6 Exercises
  5. 4 Simply connected domains
    1. 4.1 Contour integration and Cauchy’s Theorem
    2. 4.2 Green’s function
    3. 4.3 Conformal mapping and the Reimann mapping theorem
    4. 4.4 Schwarz-Christoffel transformations
    5. 4.5 Exercises
  6. 5 The reflection principle and consequences
    1. 5.1 The reflection principle for harmonic functions
    2. 5.2 Picard’s Theorem
    3. 5.3 Laurent series
    4. 5.4 Exercises
  7. 6 Boundary behavior of conformal maps
    1. 6.1 The Three Lines Lemma and integral inequalities
    2. 6.2 Carathéodory’s Theorem for Jordan domains
    3. 6.3 Exercises
  8. 7 Exit times and the growth of analytic functions
    1. 7.1 The expected exit time $T_{W}$ of Brownian motion
    2. 7.2 Unbounded domains
    3. 7.3 Moments of $T_{W}$
    4. 7.4 Exercises
  9. 8 Aspects of Stochastic Loewner Evolution
    1. 8.1 Schwarz’s lemma and consequences
    2. 8.2 de Branges’ theorem and deterministic Loewner evolution
    3. 8.3 Stochastic Loewner evolution
  10. 9 Further reading

##  Chapter 0 Preface

This was originally intended as a graduate level textbook on the connections
between planar Brownian motion and complex analysis. Then back problems forced
my retirement as a writer. I still have hope that this partial draft may be of
use to someone, somewhere, which is why I have placed it online. I am deeply
indebted to Maher Boudabra and Andrew Buttigieg for their helpful comments,
enthusiasm, and proofreading. The book, such as it is, is dedicated to my ever
patient Juyoun, without whom it would never have been written.

##  Chapter 1 Preliminaries

###  1.1 Basic notions from probability theory

We begin with a few basic notions from probability theory, and define Brownian
motion. This chapter is largely meant for review and in order to fix notation;
a reader unfamiliar with the topics discussed here is advised to consult an
introductory probability textbook. A probability space is a triple
$(\Omega,{\cal F},P)$. Here $\Omega$ is any set whatsoever; ${\cal F}$ is a
$\sigma$-field: a collection of subsets of $\Omega$, containing $\Omega$
itself, and which is closed under complements, finite intersections, and
countable unions; $P$ is a probability measure. To say that $X$ is a random
variable (abbreviated r.v.) on a probability space $(\Omega,{\cal F},P)$ means
that $X=X(\omega)$ is a function on $\Omega$, with $X^{-1}(A)\in{\cal F}$ for
all Borel sets $A$. We will also say $X$ is ${\cal F}$-measurable. If $X$ is a
random variable, there will be a unique smallest $\sigma$-field upon which $X$
is measurable; this is referred to as the $\sigma$-field generated by $X$, and
is denoted $\sigma(X)$. Two $\sigma$-fields ${\cal F}$ and ${\cal G}$ are
independent if $P(A\cap B)=P(A)P(B)$ for all $A\in{\cal F},B\in{\cal G}$. Two
random variables $X,Y$ are independent if the $\sigma$-fields
$\sigma(X),\sigma(Y)$ are independent.

Suppose that $X$ is a r.v. taking values in ${\mathbb{R}}^{n}$. If there is a
function $f_{X}(t)$ on ${\mathbb{R}}^{n}$ such that, for any Borel set
$A\subseteq{\mathbb{R}}^{n}$, we have

(1.1) $P(X\in A)=\int_{A}f_{X}(t)dt,$

then we will call $f(x)$ the probability density function (abbreviated as
p.d.f) for $X$. If $g$ is a measurable function on ${\mathbb{R}}^{n}$, then we
define the expectation of $g(X)$ to be

(1.2) $E[g(X)]=\int_{{\mathbb{R}}^{n}}g(t)f_{X}(t)dt.$

If $X$ takes values on ${\mathbb{R}}$, the mean of $X$ is then $E[X]$, and the
variance is $\mbox{Var}(X)=E[X^{2}]-E[X]^{2}$. If $X$ and $Y$ are independent,
then $E[XY]=E[X]E[Y]$ and $\mbox{Var}(X+Y)=\mbox{Var}(X)+\mbox{Var}(Y)$. The
normal or Gaussian distribution with mean $\mu$ and variance $\sigma^{2}$ is a
distribution on ${\mathbb{R}}$ characterized by the p.d.f.

(1.3)
$\rho(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}.$

We will use the notation $X\sim N(\mu,\sigma^{2})$ to signify that $X$ is a
normally distributed r.v. with mean $\mu$ and variance $\sigma^{2}$.

A stochastic process is an indexed family of r.v.’s $\\{X_{t}:t\geq 0\\}$ and
an increasing collection of $\sigma$-fields $\\{{\cal F}_{t}:t\geq 0\\}$
(which means that ${\cal F}_{s}\subseteq{\cal F}_{t}$ whenever $s\leq t$) so
that each $X_{t}$ is ${\cal F}_{t}$-measurable. The collection $\\{{\cal
F}_{t}\\}$ is called the filtration of the process, and in some sense
represents the information available to an observer at any time. In general,
we take ${\cal F}_{t}$ to be the $\sigma$-field generated by $\\{X_{s}:0\leq
s\leq t\\}$.

###  1.2 Brownian motion

A stochastic process $X_{t}$ represents a system of some sort changing over
time. Brownian motion is the stochastic process which can most aptly be
described as ”purely random motion”. That is, it represents a continuous
motion with no memory or momentum. In probabilistic terms, this requires
$X_{t}-X_{s}$ to be independent of $X_{s}$ for $s<t$: the process should have
no memory of what it did to get to $X_{s}$ in order to decide what to do to
get to $X_{t}$. The process should also move in a uniform way over different
time intervals; that is, $X_{t}-X_{s}$ should have the same distribution as
$X_{t-s}-X_{0}$, for instance. However, if these properties hold then we see
that we can write
$X_{t}-X_{0}=(X_{t/n}-X_{0})+(X_{2t/n}-X_{t/n})+\ldots(X_{nt/n}-X_{(n-1)t/n})$
for any $n\in{\mathbb{N}}$ and thereby express $X_{t}$ as the sum of a large
number of i.i.d. random variables. This brings us to the realm of the Central
Limit Theorem, which states that the sum of a large number of i.i.d. random
variables with variances will be approximately normally distributed. It is
therefore natural (and necessary) that our Brownian motion be normally
distributed.

One dimensional Brownian motion $B_{t}$ is defined to be a stochastic process
with the following properties.

  * (i)

$B_{t}:[0,\infty)\longrightarrow\mathbb{R}$ is continuous a.s.

  * (ii)

For $t_{1}<\ldots<t_{n}$ the random variables
$B_{t_{1}},B_{t_{2}}-B_{t_{1}},\ldots,B_{t_{n}}-B_{t_{n-1}}$ are independent.

  * (iii)

For $t^{\prime}<t^{\prime\prime}$, $B_{t^{\prime\prime}}-B_{t^{\prime}}$ is a
normally distributed variable with mean 0 and variance
$t^{\prime\prime}-t^{\prime}$.

In general, we will assume that $B_{0}=a$ a.s. for some constant $a$, and we
will use the notations $P_{a}$ and $E_{a}$ to mean probability and expectation
taken with respect to this initial condition. That Brownian motion even exists
is nontrivial, but well-established, and its study is a fascinating and rich
endeavor practiced by many. In this text we will focus on properties of
Brownian motion relevant to complex analysis, but so that the reader may get a
better feel for the process we briefly describe the trajectories (the image of
$B_{t}$ for a fixed $\omega$). With probability 1, Brownian paths are
nondifferentiable for every $t$ and unbounded but recurrent, which means that
they return infinitely often to every point. Brownian motion satisfies self-
similarity, which means roughly that the path looks identical regardless of
the scale at which it is viewed; this can be seen by noting that
$\frac{1}{c}B_{c^{2}t}$ for $c\in(0,\infty)$ satisfies the requirements above
and is therefore a Brownian motion. The process is therefore revealed as
possessing non-smooth, crinkled-up paths which travel slowly throughout
${\mathbb{R}}$ but always return to their starting point.

A stopping time $\tau$ is a random variable taking values in $[0,\infty]$ such
that

(1.4) $\\{\tau\leq t\\}\in{\cal F}_{t}.$

The most commonly used stopping times are hitting times, and we will make
extensive use of them in this book. For a closed set $K$ in ${\mathbb{C}}$,
let $T_{K}=\inf\\{t\geq 0:B_{t}\in K\\}$. $T_{K}$ is the hitting time of $K$,
and is a stopping time. In fact, strictly speaking the set $K$ need not be
closed in order for $T_{K}$ to be a stopping time; however, we choose to
require this because for an open set $\Omega$ we define
$T_{\Omega}=T_{\delta\Omega}=T_{\Omega^{c}}$, the hitting time of the
complement of $\Omega$ (or the exit time of $\Omega$). This is a bit of an
abuse of notation; however, the point is that we will be generally running
Brownian motion on open sets and stopping it when it exits the set or hits a
particular closed set, and we tailor our definitions accordingly.

The importance of stopping times can be seen in the following theorem, as well
as in Itō’s Theorem in the next chapter.

######  Theorem 1 (Strong Markov property)

If $\tau$ is a finite stopping time, and $\hat{B}_{t}=B_{\tau+t}-B_{\tau}$,
then $\hat{B}_{t}$ is a Brownian motion independent of $\\{B_{t}:0\leq
t\leq\tau\\}$.

The proof of this theorem is also nontrivial. The reader interested in seeing
the finer details of the result, of the construction of Brownian motion, or of
any of the other properties given here is referred to LABEL:rogers as well as
any number of other excellent texts on the subject available in any academic
library.

The primary object of interest in this book will be planar Brownian motion,
which is simply the process $(R_{t},I_{t})$ in ${\mathbb{R}}^{2}$, where
$R_{t}$ and $I_{t}$ are independent one dimensional Brownian motions.
Henceforth, any reference to Brownian motion will be to the planar variety,
unless otherwise specified. The following important observation leads to the
development in the rest of the book.

###### Theorem 2

  * (i)

Brownian motion is rotation invariant. That is, if $\Theta$ is a rotation of
the plane, then $Y_{t}=\Theta(B_{t})$ is again a Brownian motion.

  * (ii)

If $c\in[0,\infty)$, then
$Z_{t}=\frac{1}{\sqrt{c}}B_{ct}=(\frac{1}{\sqrt{c}}R_{ct},\frac{1}{\sqrt{c}}I_{ct})$
is again a Brownian motion.

Proof: That $Y_{t}$ and $Z_{t}$ are continuous with independent increments are
clear. It is also clear that the components of $Z_{t}$ are independent, and
each component can be shown to be normally distributed with the correct mean
and variance by a simple calculation. Since the components of Brownian motion
are independent, the p.d.f. for $B_{t}$ starting at 0 is simply the product of
the p.d.f.’s of the components, namely
$\rho_{t}(x,y)=\Big{(}\frac{1}{\sqrt{2\pi
t}}e^{-x^{2}/2t}\Big{)}\Big{(}\frac{1}{\sqrt{2\pi
t}}e^{-y^{2}/2t}\Big{)}=\frac{1}{2\pi t}e^{-(x^{2}+y^{2})/2t}$. Converting to
polar coordinates, this is $\rho_{t}(r,\theta)=\frac{1}{2\pi t}e^{-r^{2}/2t}$,
which is rotation invariant (i.e. does not depend on $\theta$). Thus,
$\Theta(B_{t})$ has the same distribution as $B_{t}$, namely a bivariate
normal with independent components which are normalized to have mean $0$ and
variance $t$. A similar argument holds if the Brownian motion starts at a
point other than 0.

Let us examine this theorem more closely. A rotation by $\theta$ takes the
point $(r,\theta_{0})$ in polar coordinates to $(r,\theta_{0}+\theta)$, or in
rectangular coordinates takes $(r\cos\theta_{0},r\sin\theta_{0})$ to
$(r\cos(\theta_{0}+\theta),r\sin(\theta_{0}+\theta))$. Using the angle sum
formulas for $\cos,\sin$, this is
$(r\cos\theta_{0}\cos\theta-r\sin\theta_{0}\sin\theta,r\sin\theta_{0}\cos\theta+r\cos\theta_{0}\sin\theta)$.
If we then multiply both components by $c\in{\mathbb{R}}^{+}$, we arrive at
the point $(rc\cos\theta_{0}\cos\theta-
rc\sin\theta_{0}\sin\theta,rc\sin\theta_{0}\cos\theta+rc\cos\theta_{0}\sin\theta)$,
and we know that this composition of maps takes Brownian motion to Brownian
motion (with the clock being changed according to Theorem 2). Associating this
map with the point $(c\cos\theta,c\sin\theta)$ in ${\mathbb{R}}^{2}$, we are
led naturally to the operation
$(x,y)\dot{(}x^{\prime},y^{\prime})=(xx^{\prime}-yy^{\prime},xy^{\prime}+x^{\prime}y)$,
and we will call this complex multiplication. Associating the point $(1,0)$
with the real number 1, and using the symbol $i$ for $(0,1)$, we can write any
vector in ${\mathbb{R}}^{2}$ as $x+yi$, and complex multiplication becomes
$(x+yi)(x^{\prime}+y^{\prime}i)=(xx^{\prime}-yy^{\prime})+(xy^{\prime}+x^{\prime}y)i$.
Note that $i^{2}=-1$, which is often taken as the starting point of an
investigation into complex numbers. We will refer to ${\mathbb{R}}^{2}$ with
this multiplication as the complex plane, and write ${\mathbb{C}}$ to denote
it.

###  1.3 Complex numbers

Let $z=x+yi\in{\mathbb{C}}$. We will refer to the real numbers $x$ and $y$ as
the real and imaginary parts of $z$, and write $Re(z)=x,Im(z)=y$. Define the
modulus of $z$ to be $|z|=\sqrt{x^{2}+y^{2}}$, the ordinary Euclidean metric.
We define $\bar{z}=x-iy$ and refer to $\bar{z}$ as the conjugate of z; one of
the primarily uses of $\bar{z}$ lies in the identity $z\bar{z}=|z|^{2}$. This
allows us to define quotients on ${\mathbb{C}}$ by reducing to division by
real numbers, since if $w\neq 0$ then we may write

(1.5)
$\frac{z}{w}=\frac{z\bar{w}}{|w|^{2}}=\frac{Re(z\bar{w})}{|w|^{2}}+i\frac{Im(z\bar{w})}{|w|^{2}}.$

Armed with these definitions, it is straightforward to verify the following
important facts.

###### Proposition 1

  * (i)

$|zw|=|z||w|$.

  * (ii)

$|\frac{z}{w}|=\frac{|z|}{|w|}$.

  * (iii)

(Triangle inequality) $|z+w|\leq|z|+|w|$.

  * (iv)

(Reverse triangle inequality) $|z\pm w|\geq||z|-|w||$.

  * (v)

$|z\pm w|^{2}=|z|^{2}\pm 2Re(z\bar{w})+|w|^{2}$.

The proof of this proposition is left to the reader. Fundamental to the study
of the complex numbers is Euler’s formula:

(1.6) $e^{i\theta}=\cos\theta+i\sin\theta.$

This can be verified by considering the well-known Taylor series for the real-
valued functions $e^{x},\cos x,\sin x$. This allows us to extend the domain of
the exponential function to the entire complex plane by $e^{x+yi}=e^{x}(\cos
y+i\sin y)$. The exponential so extended satisfies virtually every property
one would like; the following proposition lists some of the more basic ones.

###### Proposition 2

  * (i)

If $z=x+yi$, then $|e^{z}|=e^{x}$, and $\frac{Im(e^{z})}{Re(e^{z})}=\tan y$.

  * (ii)

The function $e^{z}$ is periodic with period $2\pi i$. That is,
$e^{z}=e^{z+2\pi i}$.

  * (iii)

$e^{z_{1}+z_{2}}=e^{z_{1}}e^{z_{2}}$.

  * (iv)

$e^{\bar{z}}=\stackrel{{\scriptstyle\line(1,0){10.0}}}{{e^{z}}}$.

Again, we leave the proof to the reader. Part $(i)$ includes an important new
fact, that the angle which the vector $e^{z}$ makes with the positive real
axis is controlled by $y$. We will define the argument of any
$z\in{\mathbb{C}}$ to be $\mbox{arg }(z)=\tan^{-1}\frac{Im(z)}{Re(z)}$; this
is clearly an ambiguous definition, since the value of $\tan^{-1}$ is only
defined up to multiples of $2\pi$, and so for exactitude we define the
principal argument of $z$ to be $\mbox{Arg }(z)=\mbox{Tan
}^{-1}\frac{Im(z)}{Re(z)}$, where here $\mbox{Tan }^{-1}$ is the branch of
$\tan^{-1}$ which takes values in $(-\pi,\pi]$. We see that $\mbox{Arg }z$ is
well defined for all $z\neq 0$, although the function $\mbox{Arg }z$ is
discontinuous on the negative real axis. We see that for any $z=x+yi\neq 0$ we
can write $z=re^{i\theta}$, where $r=|z|$ and $\theta=\mbox{Arg }z$ (or, for
that matter, any branch of $\arg z$). We will consider this to be the polar
form of $z$. Note that the properties in Propositions 1 and 2 show that
$(r_{1}e^{i\theta_{1}})(r_{2}e^{i\theta_{2}})=r_{1}r_{2}e^{i(\theta_{1}+\theta_{2})}$.

Let us now consider a number of simple transformations of ${\mathbb{C}}$. We
have seen that a rotation of the plane can be realized by multiplication by a
complex number with modulus 1, which can be expressed as $e^{i\theta}$ for
some $\theta\in{\mathbb{R}}$. That is, the function $f(z)=e^{i\theta}z$ is
simply a rotation of ${\mathbb{C}}$ by $\theta$ radians counterclockwise. By
an expansion of ${\mathbb{C}}$, we mean a function of the form $f(z)=rz$,
where $r\in[0,\infty)$; this fixes the arguments of all points in
${\mathbb{C}}$, and simply scales the moduli of all points by the
multiplicative factor $r$. A dilation is the composition of a rotation and
expansion, and can therefore be expressed as $f(z)=re^{i\theta}z$. A
translation will be a map of the form $f(z)=z+c$, which simply shifts the
entire plane by $c$. Composing a translation with a dilation gives a general
linear function $f(z)=az+c$. These are essentially the most basic functions
from ${\mathbb{C}}$ to itself, and it would be well for the reader to entirely
familiarize themselves with the geometric properties of these transformations,
as they will be important throughout.

The stereographic projection is a method of adding a ”point at infinity” to
${\mathbb{C}}$ in order to make ${\mathbb{C}}$ compact, which helps greatly in
visualizing various transformations 11endnote: 1It is also common to perform
the stereographic projection with the sphere tangent to the complex plane, for
instance by taking the sphere $\\{x_{1}^{2}+x_{2}^{2}+(x_{3}-1)^{2}=1\\}$
rather than $\SS$, or by embedding ${\mathbb{C}}$ in ${\mathbb{R}}^{2}$ as
$\\{x_{3}=-1\\}$. We prefer the setup given in the text, however, as the
important sets ${\mathbb{D}}=\\{|z|<1\\}$, $\delta{\mathbb{D}}$, and
$\\{|z|>1\\}$ appear naturally on the sphere as the intersections of $\SS$
with $\\{x_{3}<0\\},\\{x_{3}=0\\},$ and $\\{x_{3}>0\\}$, respectively. .
Imagine ${\mathbb{C}}$ embedded in
${\mathbb{R}}^{3}=\\{(x_{1},x_{2},x_{3})\\}$ as the plane $x_{3}=0$, and
consider also the unit sphere $\SS=\\{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1\\}$.
Associate any point $z=(x,y,0)\in{\mathbb{C}}$ with the point of intersection
of $\SS$ and the line passing through $(0,0,1)$ and $z$. We will let $\Theta$
denote this map from $\SS\backslash\\{(0,0,1)\\}$ to ${\mathbb{C}}$, which is
in fact a bijection. This picture may help visualize the projection.

In this way, ${\mathbb{C}}$ is essentially wrapped onto the surface of a
sphere, with only the point $(0,0,1)$ not matched to a corresponding point in
${\mathbb{C}}$. $(0,0,1)$ therefore plays a special role in this construction,
and is considered to be the point at $\infty$. Note that this is an
appropriate label, since as $|z|\longrightarrow\infty$ in ${\mathbb{C}}$,
$\Theta^{-1}(z)\longrightarrow(0,0,1)$ on $\SS$. The complex plane
${\mathbb{C}}$ together with the point at $\infty$ is commonly referred to as
the Riemann sphere, and denoted $\hat{\mathbb{C}}$. The line which passes
through $(0,0,1)$ and $z=(x,y,0)$ can be realized by $\gamma(t)=(xt,yt,1-t)$,
and this intersects $\SS$ when $x^{2}t^{2}+y^{2}t^{2}+(1-t)^{2}=1$, which
happens precisely when $(x^{2}+y^{2}+1)t^{2}-2t=0$. $t=0$ is a solution to
this quadratic equation, which yields the point of projection $(0,0,1)$. This
means that the point $\Theta^{-1}(z)$ is given by the other solution, which
occurs when $t=\frac{2}{x^{2}+y^{2}+1}$. Thus,
$\Theta^{-1}(z)=(\frac{2x}{x^{2}+y^{2}+1},\frac{2y}{x^{2}+y^{2}+1},\frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+1})$.
This formula allows us to deduce the following important fact.

###### Proposition 3

If ${\cal C}$ is a circle on $\SS$, then $\Theta({\cal C})$ is either a circle
or line in ${\mathbb{C}}$. To be specific, if ${\cal C}$ passes through
$(0,0,1)$ then $\Theta({\cal C})$ is a straight line in ${\mathbb{C}}$, while
if ${\cal C}$ does not pass through $(0,0,1)$ then $\Theta({\cal C})$ is a
circle in ${\mathbb{C}}$.

Proof: Several proofs exist; an elegant geometric one which might have pleased
the ancient Greeks is outlined in Exercise 1.4 below. However, we can give a
simple analytic one as follows. A circle ${\cal C}$ on $\SS$ can be realized
as the intersection of $\SS$ with a plane with equation
$ax_{1}+bx_{2}+cx_{3}=l$, where $a^{2}+b^{2}+c^{2}=1$ and $l\in(-1,1)$.
Substituting the formula for $\Theta^{-1}(z)$ given above into this equation
and simplifying leads to

(1.7) $(c-l)(x^{2}+y^{2})+2ax+2by=c+l.$

If $c\neq l$, then this is the equation of a circle in ${\mathbb{C}}$ (the
bijective nature of $\Theta$ shows that the circle does not degenerate). If,
however, $c=l$, then it is the equation of a line, and note in this case that
the point $(0,0,1)$ lies on ${\mathbb{C}}$.

A set $U\subseteq{\mathbb{C}}$ is open if for any $z\in U$ there is a small
disk centered at $z$ contained in $U$. A set is closed if its complement is
open. The boundary of $U$, denoted $\delta U$, is the set of all points $z$
for which there is a sequence $z_{n}$ in $U$ converging to $z$ and also a
sequence $w_{n}$ in $U^{c}$ converging to $z$. A set $U$ is connected if, for
any $z,w\in U$ there is a path in $U$ connecting them. Equivalently, an open
set is connected if it cannot be expressed as the union of two nonempty,
disjoint, open sets. A domain is an open, connected subset of ${\mathbb{C}}$.
The generic picture of a domain therefore looks something like

Our work will take place predominantly on domains: we will generally be
concerned with functions defined upon domains, and will run Brownian motions
on domains. Several domains are prominent enough to be labeled: the entire
plane ${\mathbb{C}}$; the unit disk ${\mathbb{D}}=\\{|z|<1\\}$; the upper
half-plane ${\mathbb{H}}=\\{Im(z)>0\\}$. We will also use the notation
$r{\mathbb{D}}$ to denote $\\{|z|<r\\}$.

###  1.4 Exercises

1 Verify that the p.d.f. given for normal r.v.’s in the text really is a
p.d.f.; that is, show that

(1.8)
$\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}dx=1.$

Show that if $X$ has this p.d.f., then indeed $E[X]=\mu$ and
$\mbox{Var}(X)=\sigma^{2}$. (Hint: For the first part, consider
$(\int_{-\infty}^{\infty}e^{-x^{2}}dx)(\int_{-\infty}^{\infty}e^{-y^{2}}dy)$
as an integral over the plane ${\mathbb{R}}^{2}$, and convert to polar
coordinates. For the third, make use of the first and integration by parts.)

2 Let $X$ be an $N(0,1)$ random variable. Show that

(1.9) $E[X^{n}]=\left\\{\begin{array}[]{ll}0&\qquad\mbox{if $n>0$ is odd}\\\
\frac{(2m)!}{2^{m}m!}&\qquad\mbox{if $n=2m\geq 0$ is even(Recall
$0!=1$)}\end{array}\right.$

by applying integration by parts and induction to the integral

(1.10)
$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^{n}e^{-\frac{x^{2}}{2}}dx.$

3 Let $B_{t}$ be a one dimensional Brownian motion. Show that the following
processes are also Brownian motions on $[0,T]$, where $T\in(0,\infty)$.

a) $X_{t}=-B_{t}$

b) $X_{t}=B_{T}-B_{T-t}$

c) $X_{t}=\frac{1}{c}B_{c^{2}t}$

d) $X_{t}=tB_{1/t}$ for $t>0$, $X_{0}=0$ (showing continuity of paths at $t=0$
here is difficult, so it may be assumed).

4 Let $B_{t}$ be a one dimensional Brownian motion. If $t,s>0$, show that
$E[B_{t}B_{s}]=\min(t,s)$.

5 Prove the properties of the modulus listed in Proposition 1.

6 Prove the properties of the exponential listed in Proposition 2.

7 Show that if $(x_{1},x_{2},x_{3})\in\SS$ with $x_{3}\neq 1$ then
$\Theta((x_{1},x_{2},x_{3}))=(\frac{x_{1}}{1-x_{3}},\frac{x_{2}}{1-x_{3}},0)$.

8 The purpose of this exercise is to prove Proposition 3 by purely geometric
arguments.

i) Let $l$ be a straight line in ${\mathbb{C}}$. Show that under the
stereographic projection, $l$ corresponds to a circle on $\SS$ which passes
through the point at $\infty$ using the fact that, in ${\mathbb{R}}^{3}$, a
line and any point not on the line are contained in a common plane.

ii) Let $c$ be a circle in ${\mathbb{C}}$. Show that under the stereographic
projection, $c$ corresponds to a circle on $\SS$ which does not pass through
the point at $\infty$ by following the ensuing steps.

a) We will consider ${\cal C}$ to be the intersection of $\SS$ with a plane
${\cal P}$, and by rotating if necessary we can assume that the maximal
$x_{3}$ values on ${\cal C}$ occur when $x_{2}=0$. We then take the $x_{2}=0$
slice of the configuration, and obtain something like the following picture.

Show that the triangles with vertices $\\{(0,0,1),z,w\\}$ and
$\\{(0,0,1),\Theta^{-1}(z),\Theta^{-1}(w)\\}$ are similar.

b) Conclude that the elliptical cone passing through ${\cal C}$ is intersected
by ${\cal P}$ at the same angle as by ${\mathbb{C}}=\\{x_{3}=0\\}$. Conclude
that the image under $\Theta$ of ${\cal C}$ is a circle in ${\mathbb{C}}$.

9 a) For nonzero $z\in{\mathbb{C}}$, show that $\mbox{Arg }z=\mbox{Arg
}\frac{1}{\bar{z}}$.

b) Show that the map $f(z)=\frac{1}{\bar{z}}$ with $f(0)=\infty$ and
$f(\infty)=0$ on ${\mathbb{C}}$ corresponds to a reflection over the plane
$\\{x_{3}=0\\}$ on the Riemann sphere.

###### Notes

  1. 1 It is also common to perform the stereographic projection with the sphere tangent to the complex plane, for instance by taking the sphere $\\{x_{1}^{2}+x_{2}^{2}+(x_{3}-1)^{2}=1\\}$ rather than $\SS$, or by embedding ${\mathbb{C}}$ in ${\mathbb{R}}^{2}$ as $\\{x_{3}=-1\\}$. We prefer the setup given in the text, however, as the important sets ${\mathbb{D}}=\\{|z|<1\\}$, $\delta{\mathbb{D}}$, and $\\{|z|>1\\}$ appear naturally on the sphere as the intersections of $\SS$ with $\\{x_{3}<0\\},\\{x_{3}=0\\},$ and $\\{x_{3}>0\\}$, respectively.

##  Chapter 2 Dynkin’s formula and Lévy’s Theorem

In this chapter we begin with the formula which governs the manner at which
the expectation of $u(B_{t})$ evolves over time, where $u$ is any $C^{2}$
function on a domain in ${\mathbb{C}}$. We will see that this formula, which
we will refer to as Dynkin’s formula, leads naturally to the definitions of
harmonic and analytic functions and the power series expansion of each. It may
therefore be considered one of the fundamental formulas in complex analysis,
in addition to its obvious importance in probability theory.

###  2.1 Dynkin’s formula

The Laplacian differential operator $\Delta$ is defined by
$\Delta=\frac{d^{2}}{dx^{2}}+\frac{d^{2}}{dy^{2}}$. This can be applied to a
real-valued function on a domain in ${\mathbb{C}}$, and if $u$ is complex-
valued, so that $u=u_{1}+iu_{2}$ with $u_{1},u_{2}$ real-valued, then we
simply have $\Delta u=(\Delta u_{1})+i(\Delta u_{2})$. The following theorem
indicates the prominent position of the Laplacian in the study of Brownian
motion.

###### Theorem 3

Suppose that $u(z)=u(x,y)$ is a $C^{2}$ function on a bounded domain
$\Omega\subseteq{\mathbb{C}}$, and let $a\in\Omega$. Suppose $\tau<T_{\Omega}$
with $E[\tau]<\infty$ is a stopping time such that $u$ and its derivatives up
to order 2 are a.s. bounded on the set $\\{B_{t}:0\leq t\leq\tau,B_{0}=a\mbox{
a.s.}\\}$. Then

(2.1) $E_{a}[u(B_{\tau})]-u(a)=\frac{1}{2}E_{a}\int_{0}^{\tau}\Delta
u(B_{s})ds.$

If $u$ extends continuously to $\delta\Omega$, then the same conclusion holds
even if we relax the requirement $\tau<T_{\Omega}$ to $\tau\leq T_{\Omega}$.

Proof: By considering real and imaginary parts separately, we may assume $u$
is real valued. We begin by partitioning $[0,\infty)$ as
$0=s_{0}<s_{1}<s_{2}<\ldots$ with $s_{n}\nearrow\infty$, and write using
independence

(2.2)
$\begin{split}E_{a}[u(B_{\tau})]-u(a)&=\sum_{n=0}^{\infty}E_{a}[u(B_{s_{n+1}\wedge\tau})-u(B_{s_{n}\wedge\tau})]\\\
&\approx\sum_{n=0}^{\infty}E_{a}[(u(B_{s_{n+1}})-u(B_{s_{n}}))1_{\\{\tau\geq
s_{n}\\}}]\end{split}$

Taylor’s Theorem allows us to write

(2.3) $\displaystyle\hskip 28.45274ptu(B_{s_{n+1}})-u(B_{s_{n}})=$
$\displaystyle
u_{x}(B_{s_{n}})(R_{s_{n+1}}-R_{s_{n}})+u_{y}(B_{s_{n}})(I_{s_{n+1}}-I_{s_{n}})+\frac{1}{2}u_{xx}(B_{s_{n}})(R_{s_{n+1}}-R_{s_{n}})^{2}$
$\displaystyle+u_{xy}(B_{s_{n}})(R_{s_{n+1}}-R_{s_{n}})(I_{s_{n+1}}-I_{s_{n}})+\frac{1}{2}u_{yy}(B_{s_{n}})(I_{s_{n+1}}-I_{s_{n}})^{2}+o(|B_{s_{n+1}}-B_{s_{n}}|^{2}),$

where as before $R$ and $I$ are the real and imaginary parts of $B$,
respectively. Note that $(R_{s_{n+1}}-R_{s_{n}})$ and
$(I_{s_{n+1}}-I_{s_{n}})$ are independent of $1_{\\{\tau\geq s_{n}\\}}$, since
$\tau$ is a stopping time, and also independent of the terms of the form
$u_{x}(B_{s_{n}}),u_{yy}(B_{s_{n}})$, etc. Thus, when we plug (2.3) into (2.2)
we may factor the expectation and use the facts that
$E_{a}[R_{s_{n+1}}-R_{s_{n}}]=E_{a}[I_{s_{n+1}}-I_{s_{n}}]=E_{a}[(R_{s_{n+1}}-R_{s_{n}})(I_{s_{n+1}}-I_{s_{n}})]=0$
and
$E_{a}[(R_{s_{n+1}}-R_{s_{n}})^{2}]=E_{a}[(I_{s_{n+1}}-I_{s_{n}})^{2}]=\frac{1}{2}E_{a}[|B_{s_{n+1}}-B_{s_{n}}|^{2}]=(s_{n+1}-s_{n})$.
We arrive at

(2.4)
$E_{a}[u(B_{\tau})]-u(a)=\frac{1}{2}\sum_{n=0}^{\infty}E_{a}[(u_{xx}(B_{s_{n}})+u_{yy}(B_{s_{n}}))1_{\\{\tau\geq
s_{n}\\}}](s_{n+1}-s_{n})+\sum_{n=0}^{\infty}o(s_{n+1}-s_{n})E_{a}[1_{\tau\geq
s_{n}}].$

We now let the partition become finer, so that the mesh size goes to $0$.
Since $E_{a}[\tau]<\infty$, the last term approaches 0, and the first term
approaches $\frac{1}{2}E_{a}\int_{0}^{\tau}\Delta u(B_{s})ds$, giving (2.1).
We made use of a number of approximations along the way, but since $u$ and its
derivatives are bounded in the interior of $\gamma$ these approximations
behave correctly in the limit, and the result follows.

Remark: The boundedness assumptions on $u$ and $\Omega$ are important; see
Exercise 2.5. The requirement that $\tau$ must be a stopping time is also
important; see Exercise 2.5.

The conditions on the stopping time in the statement of the theorem may appear
to be difficult to check in practice. However, we will use primarily one type
of stopping time in order to apply this theorem: the hitting time of a
bounded, closed curve $\gamma$ which lies in $\Omega$ and separates $a$ from
$\delta\Omega$.

The fact that $u$ is $C^{2}$ implies easily that $T_{\gamma}$ automatically
satisfies the required boundedness condition on $u$ and its derivatives, so
that it must only be shown that $E[T_{\gamma}]<\infty$. Since any bounded
curve is contained in a disk centered at $a$, this is an immediate consequence
of the following proposition.

###### Proposition 4

$E_{0}[T_{r{\mathbb{D}}}]=\frac{r^{2}}{2}$.

Proof: Fix large $N>0$ and let $\tau=\tau(N)=T_{r{\mathbb{D}}}\wedge N$. Note
that $E[\tau]<N<\infty$, so we may apply Dynkin’s formula with
$u(x+yi)=x^{2}+y^{2}$. We have $u(0)=0$ and $\Delta u\equiv 4$; we therefore
obtain

(2.5) $E_{0}[u(B_{\tau})]=\frac{1}{2}E_{0}\int_{0}^{\tau}4ds=2E_{0}[\tau].$

Since $u\leq r^{2}$ on $\\{B_{t}:0\leq t\leq\tau\\}$, we see that $E[\tau]\leq
r^{2}/2$ independently of $N$, and we may let $N\nearrow\infty$ and apply
monotone convergence to conclude that $E[T_{r{\mathbb{D}}}]<\infty$. This
implies that $T_{r{\mathbb{D}}}<\infty$ a.s. so that
$E_{0}[u(B_{T_{r{\mathbb{D}}}})]=r^{2}$, and also allows us to apply Dynkin’s
formula directly to $T_{r{\mathbb{D}}}$ in order to conclude that
$r^{2}=2E_{0}[T_{r{\mathbb{D}}}]$.

We now have a large class of stopping times for which we can apply Dynkin’s
formula. However, for arbitrary $u$ it is of course difficult to say much
about the integral term $\int_{0}^{\tau}\Delta u(B_{s})ds$; nevertheless,
there is one instance in which this expression is easy to determine, and this
leads to an immensely valuable definition. We will say that the function $u$
is harmonic if $\Delta u(z)=0$ for all $z$ in the domain of $u$. It is then
immediate from Theorem 3 that $E_{a}[u(B_{\tau})]=u(a)$. Taking
$u(x+yi)=x+yi$, which is clearly harmonic, gives

###### Corollary 2

Suppose $u$ is harmonic and $\tau$ is a stopping time satisfying the
conditions of Theorem 3. Then $E_{a}[u(B_{\tau})]=u(a)$.

In particular, if we apply this corollary to the identity function $u(z)=z$,
which is trivially harmonic, we see that if $E[\tau]<\infty$ is a stopping
time such that $\sup\\{|B_{t}|:0\leq t\leq\tau,B_{0}=a\mbox{ a.s.}\\}<\infty$
then $E_{a}[B_{\tau}]=a$. This fact admits a fruitful interpretation. If $f$
is a function such that $f(B_{t})$ is a Brownian motion, then we will have
$E_{a}[f(B_{\tau})]=f(a)$. This will hold for a Brownian motion run at a
different speed, since all that matters is the values of $B_{\tau}$, and
therefore by Proposition 2 will hold for any linear function $f(z)=cz+b$; but
since $\tau$ is a variable time it will also hold for a Brownian motion run at
a variable speed. In order to make this precise, we define a time change to be
a family of stopping times $\\{C_{t}:t\geq 0\\}$ with respect to a given
filtration which is a.s. non-decreasing and right continuous in $s$. We will
then say that a process $X_{t}$ is a time changed Brownian motion if there is
a time change $C_{t}$ of $X_{t}$ such that $X_{C_{t}}$ is a Brownian motion;
or, equivalently, there is a Brownian motion $B_{t}$ and a time change $C_{t}$
of $B_{t}$ such that $B_{C_{t}}=X_{t}$ a.s. The prior argument shows that if
we have a function $f$ such that $f(B_{t})$ is a time changed Brownian motion,
then $E_{a}[f(B_{\tau})]=f(a)$ for all suitable stopping times $\tau$. It
should be no great stretch of the imagination to suppose that this equation is
extremely useful, though just how useful it will turn out to be may surprise
the reader. Suffice it to say that a large percentage of what follows in this
book can be reduced at its core to an application of this result. In any case,
the next logical task is to identify a large class of functions which map
Brownian motion to a time change of Brownian motion. This is the subject of
the next section.

###  2.2 Analytic invariance of Brownian motion

We have seen in Proposition 2 that Brownian motion is invariant under
dilations and translations, and therefore under linear maps $f(z)=cz+b$,
although the time must be adjusted by the square of the expanding factor
$|c|$. It turns out that this phenomenon holds for the much more general class
of functions which are ”locally dilating”, in the following sense. We will
call a function $f$ on a domain $\Omega$ analytic if it can be expressed as

(2.6) $f(z)=f(z_{0})+c(z-z_{0})+o(|z-z_{0}|)$

around any point $z_{0}\in\Omega$, where $c\in{\mathbb{C}}$ is a constant
(depending upon $z_{0}$). That is, $f$ is a continuous function whose behavior
close to $z_{0}$ is dominated by a dilation term. Equivalently, we may say
that $f(z)$ is analytic if

(2.7) $\lim_{\Delta z\longrightarrow 0}\frac{f(z_{0}+\Delta
z)-f(z_{0})}{\Delta z}$

exists for all $z_{0}\in\Omega$. In other words, analytic functions are simply
complex differentiable functions, and for that reason we will denote the limit
in (2.7) as $f^{\prime}(z_{0})$; however, the reader is forewarned that the
behavior of analytic functions is far more restricted than that of
differentiable functions of a real variable, and not to attempt to analogize
too much between these two types of differentiable functions. This will be
revealed as we progress through the subject.

The primary connection between analytic functions and Brownian motion lies in
the following beautiful result, which is known as Lévy’s Theorem.

###### Theorem 4

Let $f$ be analytic and nonconstant on a domain $U$, and let $a\in U$. Let
$B_{t}$ be a planar Brownian motion which starts at $a$ and is stopped at a
stopping time $\tau$ such that $\tau\leq T(U)$ a.s. Set

$\sigma(t)=\int_{0}^{t\wedge\tau}|f^{\prime}(B_{s})|^{2}ds.$

for $t\in[0,\tau]$. $\sigma(t)$ is a.s. strictly increasing and continuous,
and we let $C_{t}=\sigma^{-1}(t)$ on $[0,\sigma(\tau)]$. Then
$\hat{B}_{t}=f(B_{C_{t}})$ is a Brownian motion stopped at the stopping time
$\sigma(\tau)$.

We will often refer to the new Brownian motion $\hat{B}_{t}$ and stopping time
$\sigma(\tau)$ as the projections of $B_{t}$ and $\tau$ under $f$. An
intuitive proof of the theorem is simple. We know that dilations preserve
Brownian motion, but that the time scale must be adjusted by the square of the
expanding factor. Therefore, locally dilating maps preserve Brownian motion as
well, except that in the more general setting the expansion factor is
different at different points in the domain. We must therefore scale the time
of the transformed Brownian motion according to the accumulated square of the
expansion factors at all of the points that the Brownian motion has visited,
which is done via $\sigma_{t}$ and its inverse $C_{t}$. Q.E.D.

A rigorous proof of Theorem 4 would require quite a bit of work, including the
introduction of stochastic integration. Although this is a fascinating and
important subject in its own right, we choose not to include it in our
development. A reader unsatisfied with the intuitive explanation given here is
advised to consult [41], [38], or [70] for complete proofs of the result.

Note also that, for a analytic function $f$, we have
$f(a)=E_{f(a)}[B_{C_{\sigma_{T}}}]=E_{a}[f(B_{T})]$; this is the same formula
which harmonic functions satisfy, though at this stage we do not know that
analytic functions are harmonic, or even that they are $C^{2}$. As we will see
in the last section of this chapter, however, analytic functions are indeed
harmonic.

Now that we have the concept of analytic functions, a natural question arises:
what functions are analytic? We begin by noting that the standard proofs from
calculus that sums, differences, products, quotients (with nonzero
denominators), and compositions of differentiable functions are again
differentiable carry over directly into the complex setting, so that we may
build analytic functions by combining others, with the formulas for the
derivatives given by the ordinary sum, product, quotient, and chain rules. We
may therefore commence with the observations that the constant function
$f(z)=c$ and identity function $f(z)=z$ are trivially analytic, which
immediately gives us all polynomials, with the derivative of
$\sum_{n=0}^{d}a_{n}z^{n}$ given by $\sum_{n=1}^{d}na_{n}z^{n-1}$. It is
natural to ask whether we may carry this over to ”infinite degree
polynomials”; that is, if $f(z)=\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$ is a
power series, is $f(z)$ analytic? The following proposition answers this
question.

###### Proposition 5

Consider the expression $\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$. Let
$R\in[0,\infty]$ be given by
$\frac{1}{R}=\limsup_{n\longrightarrow\infty}|a_{n}|^{1/n}$.

  * (i)

The partial sums $\sum_{n=0}^{M}a_{n}(z-z_{0})^{n}$ converge on $D(z_{0},R)$
and do not converge on
$\stackrel{{\scriptstyle\line(1,0){36.0}}}{{D(z_{0},R)}}^{c}=\\{|z-z_{0}|>R\\}$
(if $R=0$, this signifies convergence only at $z=z_{0}$, and if $R=\infty$
this signifies convergence for all $z$).

  * (ii)

This convergence is uniform on $D(z_{0},r)$ for any $r<R$.

  * (iii)

If we let
$f(z)=\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}=\lim_{M\longrightarrow\infty}\sum_{n=0}^{M}a_{n}(z-z_{0})^{n}$,
then $f$ is analytic on $D(z_{0},R)$, and
$f^{\prime}(z)=\sum_{n=1}^{\infty}na_{n}(z-z_{0})^{n-1}$.

Proof: If $|z-z_{0}|<r<R$, then

(2.8)
$\Big{|}\sum_{n=N}^{M}a_{n}(z-z_{0})^{n}\Big{|}\leq\sum_{n=N}^{M}(|a_{n}|^{1/n}|z-z_{0}|)^{n}\leq\sum_{n=N}^{\infty}((\frac{1}{R}+\varepsilon)r)^{n},$

where $\varepsilon$ can be made arbitrarily small by requiring $N$ and $M$ to
be sufficiently large. We may therefore take $(\frac{1}{R}+\varepsilon)r<1$,
and it follows that $\sum_{n=N}^{M}a_{n}(z-z_{0})^{n}\longrightarrow 0$
uniformly as $N,M\longrightarrow\infty$. This implies that
$\sum_{n=0}^{M}a_{n}(z-z_{0})^{n}$ is a Cauchy sequence, uniformly on
$D(z_{0},r)$, proving $(ii)$ and the first part of $(i)$. The second part of
$(i)$ follows easily in a similar manner. To prove $(iii)$, we use the
identity $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1}$ and the
uniform convergence to calculate

(2.9) $\begin{split}f^{\prime}(z)&=\lim_{h\longrightarrow
0}\frac{\sum_{n=0}^{\infty}a_{n}(z+h-z_{0})^{n}-\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}}{h}\\\
&=\lim_{h\longrightarrow
0}((z+h-z_{0})-(z-z_{0}))\frac{\sum_{n=0}^{\infty}a_{n}((z+h-z_{0})^{n-1}+(z+h-z_{0})^{n-2}(z-z_{0})+\ldots+(z-z_{0})^{n-1})}{h}\\\
&=\lim_{h\longrightarrow
0}\sum_{n=0}^{\infty}a_{n}((z+h-z_{0})^{n-1}+(z+h-z_{0})^{n-2}(z-z_{0})+\ldots+(z-z_{0})^{n-1})\\\
&=\sum_{n=1}^{\infty}na_{n}(z-z_{0})^{n-1}.\end{split}$

We see that convergent power series give us analytic functions, so by taking
the Taylor series of well-known transcendental functions, such as the
exponential and trigonometric ones, we obtain a wealth of analytic functions.
But what about analytic functions not given by power series? It turns out that
there are none, at least in a local sense: every analytic function can be
expressed as a convergent power series over any disc in its domain. Our next
order of business will be to prove this important fact, which we will do in
Section 2.4. First, however, we must study a prominent class of analytic
functions known as Möbius transformations.

###  2.3 Möbius transformations

This section is devoted to the class of analytic functions known as Möbius
transformations. We will see that these functions allow us to transform any
three points on the plane to any other three, leading to useful
simplifications in many cases. This will lead naturally in the next section to
an expression for the exit distribution of Brownian motion from a disc, which
in turn leads to the Poisson integral formula and the expansion into power
series for harmonic functions, two of of the cornerstones of complex analysis.
Furthermore, the class of analytic self-maps of the Reimann sphere is
coincident with the class of Möbius transformations. Fluency in these maps is
therefore a must for any complex analyst.

A Möbius transformation is a function of the form $\phi(z)=\frac{az+b}{cz+d}$
for $a,b,c,d\in{\mathbb{C}},ad-bc\neq 0$. The determinant condition $ad-bc\neq
0$ is simply to ensure that $\phi(z)$ is not a constant. We have seen two
types of Möbius transformations already: the translation $\phi(z)=z+b$ and the
dilation $\phi(z)=az$. Another simple example is the inversion $\phi(z)=1/z$.
It is a standard and useful fact that these simple forms can be used to build
all other Möbius transformations.

###### Lemma 1

Any Möbius transformation can be expressed as the composition of translations,
dilations, and the inversion.

Proof: Let $\phi(z)=\frac{az+b}{cz+d}$ be an arbitrary Möbius transformation.
If $c=0$ then $\phi$ is merely a linear map, which is clearly the composition
of a translation and dilation. If $c\neq 0$, let $\phi_{1}(z)=z+d/c$,
$\phi_{2}(z)=1/z$, $\phi_{3}(z)=\frac{(bc-ad)}{c^{2}}z$, and
$\phi_{4}(z)=z+a/c$. Then the reader may check that
$\phi=\phi_{4}\circ\phi_{3}\circ\phi_{2}\circ\phi_{1}$.

If $c\neq 0$ then it may seem that $\phi$ is undefined at $z=-d/c$; however,
it is far more profitable to consider $\phi$ as a map from the Riemann sphere
to itself with the definitions $\phi(-d/c)=\infty$ and $\phi(\infty)=a/c$.
Defined as such, Möbius transformations are bijections from the Riemann sphere
to itself (this is left to the reader to check, in Exercise 2.5 below), and as
quotients of polynomials are analytic at all points as well (analyticity at
$-d/c$ means that $1/\phi$ is analytic there, and analyticity at $\infty$
means that $\phi(1/z)$ is analytic at 0). If $c=0$, so that $\phi$ is a linear
map, we define $\phi(\infty)=\infty$, and likewise obtain a analytic bijection
of the sphere to itself. Möbius transformations are most naturally understood
on the Riemann sphere rather than in the plane, as the following important
proposition indicates.

###### Proposition 6

Möbius transformations map circles to circles.

Remark: This and ensuing statements about circles should be taken in the
context of the Riemann sphere, and recall from Section 1.3 that a circle on
the sphere is a circle or straight line in the plane. Thus, the image of a
circle or straight line in the plane is either a circle or straight line in
the plane, but a circle may be mapped to a line and vice versa.

Proof: Lemma 1 shows us that we need only prove the proposition for
translations, dilations, and the inversion. We will take it as evident for
translations and dilations, and for the inversion we note that a circle on the
sphere can be realized as all points $z$ which satisfy

(2.10) $a|z|^{2}+bz+c\bar{z}+d=0,$

for some $a,b,c,d\in{\mathbb{C}}$ (a straight line in the plane occurs
precisely when $a=0$). Replacing $z$ by $\frac{1}{z}$ and multiplying through
by $|z|^{2}$ gives

(2.11) $a+b\bar{z}+cz+d|z|^{2}=0,$

again the equation of a circle on the sphere (see also Exercise 2.5 below).

###### Proposition 7

  * (i)

The composition of two Möbius transformations is again a Möbius
transformation. Furthermore, if $\phi(z)=\frac{az+b}{cz+d}$ is a Möbius
transformation, then $\phi^{-1}$ is as well, and is given by
$\phi^{-1}(w)=\frac{dw-b}{-cw+a}$.

  * (ii)

If $(a,b,c),(a^{\prime},b^{\prime},c^{\prime})$ are a pair of sets of three
distinct points on the sphere, then there is a Möbius transformation taking
$a\longrightarrow a^{\prime},b\longrightarrow b^{\prime}$, and
$c\longrightarrow c^{\prime}$.

  * (iii)

If $\phi_{1}$ and $\phi_{2}$ are Möbius transformations which agree at three
distinct points then they coincide for all $z$.

Remark: The formula for the inverse in $(i)$ may look familiar, and this is no
coincidence; see Exercise 2.5 below.

Proof: $(i)$ is a simple calculation. For $(ii)$, let
$\phi_{a,b,c}=\frac{(b-c)(z-a)}{(b-a)(z-c)}$; here
$\phi_{a,b,c}(a)=0,\phi_{a,b,c}(b)=1$, and $\phi_{a,b,c}(c)=\infty$. Define
$\phi_{a^{\prime},b^{\prime},c^{\prime}}$ accordingly. Then
$\phi_{a^{\prime},b^{\prime},c^{\prime}}^{-1}\circ\phi_{a,b,c}$ takes
$a\longrightarrow a^{\prime},b\longrightarrow b^{\prime}$, and
$c\longrightarrow c^{\prime}$. $(ii)$ allows us to pre- and postcompose by
appropriate Möbius transformations in order to reduce $(iii)$ to showing that
a Möbius transformation which fixes $0,1,$ and $\infty$ must be the identity;
however, a Möbius transformation fixing $\infty$ is simply a linear map, and
the only degree 1 polynomial with two fixed points is the identity, so the
result follows.

Proposition 6 shows that circles are important when studying Möbius
transformations, and Proposition 7 shows that triples of points are as well.
This suggests that triples of points are closely connected with circles, and
this is indeed so.

###### Proposition 8

If $a,b,c$ are distinct points on $\hat{\mathbb{C}}$, then there is precisely
one circle on the sphere which passes through all three.

Proof: If one of the points is $\infty$, then it is easy to see that the
circle must be the line in the plane through the other two. Otherwise,
construct the perpendicular bisectors of the line segments
$\stackrel{{\scriptstyle\hskip 3.61371pt\line(1,0){15.0}}}{{ab}}$ and
$\stackrel{{\scriptstyle\hskip 3.61371pt\line(1,0){15.0}}}{{bc}}$. If the two
bisectors are parallel, then $a,b,c$ are colinear, and the circle on the
sphere in question is the straight line through the three of them. If, on the
other hand, the two bisectors intersect, then this point of intersection is
the unique point in the plane equidistant from $a,b,c$, and is therefore the
center of the unique circle passing through the three of them.

This and the previous propositions lead to an easy test as to whether a circle
can be found which passes through four given points: simply find the Möbius
transformation mapping three of the points to $0,1,\infty$, and see where it
sends the fourth point. If the fourth point is sent to the real axis, then the
four points lie on a circle, but if it is not then they do not. With this end
in mind, define the cross ratio of the points $z_{1},z_{2},z_{3},z_{4}$ as

(2.12)
$(z_{1},z_{2};z_{3},z_{4})=\frac{(z_{1}-z_{3})(z_{2}-z_{4})}{(z_{2}-z_{3})(z_{1}-z_{4})}.$

The reader may compare with the formula given in Proposition 7 to see that
this is the image of $z_{4}$ under the Möbius transformation taking
$z_{1}\longrightarrow\infty$, $z_{2}\longrightarrow 0$, and
$z_{3}\longrightarrow 1$. We see that four points lie on a circle if their
cross ratio is real. Proposition 6 then implies that, for any Möbius
transformation $\phi$, $(z_{1},z_{2};z_{3},z_{4})$ is real if and only if
$(\phi(z_{1}),\phi(z_{2});\phi(z_{3}),\phi(z_{4}))$ is real. However, much
more is true.

###### Proposition 9

For any Möbius transformation $\phi$ and any four points
$z_{1},z_{2},z_{3},z_{4}$,

(2.13)
$(\phi(z_{1}),\phi(z_{2});\phi(z_{3}),\phi(z_{4}))=(z_{1},z_{2};z_{3},z_{4}).$

Proof: This is a simple calculation for translations, dilations, and the
inversion. Proposition 1 therefore gives the result.

For $a\in{\mathbb{D}}$ consider the Möbius transformation

(2.14) $\psi_{a}(z)=\frac{z-a}{1-\bar{a}z}.$

It can be shown that for $|z|\leq 1$ we have $|\psi_{a}(z)|\leq 1$, with
equality precisely when $|z|=1$ (see Exercise 2.5 below). Thus, $\psi_{a}$ is
a analytic self-map of ${\mathbb{D}}$ sending $a$ to 0. This map will allow us
to give an easy solution to an important problem. Suppose we distinguish an
interval ${\cal I}$ on the unit circle; for $a\in{\mathbb{D}}$, what is
$P_{a}(B_{T_{\mathbb{D}}}\in{\cal I})$? That is, what is the probability that
a Brownian motion starting at $a$ will exit ${\mathbb{D}}$ through ${\cal I}$?
We first note that the rotation invariance of Brownian motion shows that, if
$a=0$, then the quantity $P_{0}(B_{T_{\mathbb{D}}}\in{\cal I})$ must be a
rotation invariant measure on $\\{|z|=1\\}$; and this implies that it is equal
to $\frac{m({\cal I})}{2\pi}$, where $m({\cal I})$ is the Lebesgue measure, or
length, of ${\cal I}$. For $a\neq 0$, we may simply use the map $\psi_{a}$ to
map $a$ to $0$ and use the fact that $\psi_{a}(B_{t})$ is a time-changed
Brownian motion; note that our question is concerned with the exit probability
of the Brownian motion but not the time, and therefore we need not even bother
with keeping track of the time change. Clearly, the Brownian motion $B_{t}$
leaves ${\mathbb{D}}$ through ${\cal I}$ if, and only if, the Brownian motion
$\psi_{a}(B_{t})$ leaves ${\mathbb{D}}$ through $\psi_{a}({\cal I})$:

We obtain the following identity:

(2.15) $P_{a}(B_{T({\mathbb{D}})}\in{\cal
I})=P_{0}(B_{T({\mathbb{D}})}\in\psi_{a}({\cal I}))=\frac{m(\psi_{a}({\cal
I}))}{2\pi}.$

To calculate $m(\psi_{a}({\cal I}))$, note simply that the map $\psi_{a}$
expands an infinitesimal portion of ${\cal I}$ centered at a point
$e^{it}\in{\cal I}$ with expansion factor $|\psi_{a}^{\prime}(e^{it})|$. It
follows that

(2.16) $P_{a}(B_{T({\mathbb{D}})}\in{\cal I})=\frac{m(\psi_{a}({\cal
I}))}{2\pi}=\frac{1}{2\pi}\int_{{\cal
I}}|\psi_{a}^{\prime}(e^{it})|dt=\frac{1}{2\pi}\int_{{\cal
I}}\frac{1-|a|^{2}}{|1-\bar{a}e^{it}|^{2}}dt.$

In the next section we will explore applications of this formula for harmonic
and analytic functions.

###  2.4 Poisson’s integral formula and power series

Let us suppose that we have a harmonic function $h$ on ${\mathbb{D}}$ which is
continuous on $\stackrel{{\scriptstyle\line(1,0){10.0}}}{{{\mathbb{D}}}}$. We
then have by Dynkin’s formula, for any $a\in{\mathbb{D}}$,

(2.17)
$h(a)=E_{a}[B_{T_{\mathbb{D}}}]=\int_{0}^{2\pi}h(e^{it})P_{a}(B_{T({\mathbb{D}})}\in
dt).$

Recalling the formula for exit probabilities from the previous section, we let
${\cal I}$ shrink to an infinitesimal interval to obtain the identity

(2.18) $P_{a}(B_{T({\mathbb{D}})}\in
dt)=\frac{1}{2\pi}|\psi_{a}^{\prime}(e^{it})|dt=\frac{1}{2\pi}\frac{1-|a|^{2}}{|1-\bar{a}e^{it}|^{2}}dt.$

Let ${\cal P}_{r}(\theta)=\frac{1-r^{2}}{|1-re^{-i\theta}|^{2}}$; this is
known as the Poisson kernel. If we set $a=re^{i\theta}$, then (2.18) gives us

######  Theorem 5 (Poisson integral formula)

Let $h$ be a harmonic function on ${\mathbb{D}}$ which is continuous on
$\stackrel{{\scriptstyle\line(1,0){10.0}}}{{{\mathbb{D}}}}$. Then

(2.19) $h(re^{i\theta})=\frac{1}{2\pi}\int_{0}^{2\pi}h(e^{it}){\cal
P}_{r}(\theta-t)dt.$

There are several equivalent forms for the Poisson kernel (see Exercise 2.5
below), one of which is ${\cal
P}_{r}(\theta)=\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}$; this is an
extremely useful fact, since it allows us to write

(2.20)
$\begin{split}h(a)&=h(re^{i\theta})=\frac{1}{2\pi}\int_{0}^{2\pi}h(e^{it})\Big{(}\sum_{n=-\infty}^{\infty}r^{|n|}e^{in(\theta-t)}\Big{)}dt=\\\
&=\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}\frac{1}{2\pi}\int_{0}^{2\pi}h(e^{it})e^{-int}dt\\\
&=\beta_{0}+\sum_{n=1}^{\infty}\beta_{-n}\bar{a}^{n}+\sum_{n=1}^{\infty}\beta_{n}a^{n},\end{split}$

where

(2.21) $\beta_{n}=\frac{1}{2\pi}\int_{0}^{2\pi}h(e^{it})e^{-int}dt.$

Note that $h$ is continuous on $\bar{\mathbb{D}}$ and thus bounded there, so
that interchanging summation and integration in (2.20) is justified. We see
that $h$ can be expressed as the sum of power series in $a$ and $\bar{a}$.
There is nothing special about the unit disk for the purposes of this
argument; it applies equally well to disks with arbitrary centers and radii,
and we therefore obtain

###### Theorem 6

Let $h(z)$ be harmonic (or analytic) on $D(z_{0},R)$. Then we may write

(2.22)
$h(z)=\beta_{0}+\sum_{n=1}^{\infty}\beta_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}+\sum_{n=1}^{\infty}\beta_{n}(z-z_{0})^{n},$

for all $z\in D(z_{0},R)$, with $\beta_{n}$ given by

(2.23) $\beta_{n}=\frac{1}{2\pi
r^{n}}\int_{0}^{2\pi}h(z_{0}+re^{it})e^{-int}dt,$

for any $0<r<R$. These series converge uniformly on $D(z_{0},R^{\prime})$ for
any $R^{\prime}<R$. Furthermore, the coefficients in the series are unique,
and $h$ is real valued if and only if $\beta_{-n}=\bar{\beta}_{n}$ for all
$n$.

Proof: The first statement has already been proved, and the uniform
convergence follows as in Proposition 5. The final statement follows from a
standard technique in Fourier analysis; see Exercise 2.5 below.

We will somewhat informally refer to the first sum in (2.40) as the
$\bar{z}$-part, and the second as $z$-part, of the power series for $h(z)$.
Since $z$ and $\bar{z}$ are simply linear combinations of $x$ and $y$, we
obtain

###### Corollary 3

If $h$ is harmonic or analytic, then $h$ is $C^{\infty}$.

Define the differential operators

(2.24)
$\frac{d}{dz}=\frac{1}{2}(\frac{d}{dx}-i\frac{d}{dy}),\qquad\frac{d}{d\bar{z}}=\frac{1}{2}(\frac{d}{dx}+i\frac{d}{dy}).$

These are known as the Wirtinger derivatives, and satisfy a number of nice
properties, such as

###### Lemma 2

Suppose $f,g$ are differentiable functions, and $a,b\in{\mathbb{C}}$. Then

  * (i)

(Sum/difference rule)

(2.25)
$\frac{d}{dz}(af+bg)=a\frac{df}{dz}+b\frac{dg}{dz},\qquad\frac{d}{d\bar{z}}(af+bg)=a\frac{df}{d\bar{z}}+b\frac{dg}{d\bar{z}}.$

  * (ii)

(Product rule)

(2.26)
$\frac{d}{dz}(fg)=f\frac{dg}{dz}+g\frac{df}{dz},\qquad\frac{d}{d\bar{z}}(fg)=f\frac{dg}{d\bar{z}}+g\frac{df}{d\bar{z}}.$

  * (iii)

(Chain rule)

(2.27) $\frac{d}{dz}(f\circ g)=\Big{(}\frac{df}{dz}\circ
g\Big{)}\frac{dg}{dz}+\Big{(}\frac{df}{d\bar{z}}\circ
g\Big{)}\frac{d\bar{g}}{dz},\qquad\frac{d}{d\bar{z}}(f\circ
g)=\Big{(}\frac{df}{dz}\circ
g\Big{)}\frac{dg}{d\bar{z}}+\Big{(}\frac{df}{d\bar{z}}\circ
g\Big{)}\frac{d\bar{g}}{d\bar{z}}.$

  * (iv)

For integer $n\geq 1$, we have

(2.28)
$\frac{d}{dz}z^{n}=nz^{n-1},\qquad\frac{d}{d\bar{z}}\bar{z}^{n}=n\bar{z}^{n-1},\qquad\frac{d}{dz}\bar{z}^{n}=\frac{d}{d\bar{z}}z^{n}=0.$

  * (v)

If $h$ is $C^{2}$, then

(2.29) $\Delta h=4\frac{d^{2}h}{dzd\bar{z}}=4\frac{d^{2}h}{d\bar{z}dz}.$

The properties given in this lemma are all consequences of standard rules in
multivariate calculus, and their proofs are left to the reader (Exercise 2.5).
An immediate consequence of the lemma is the following.

###### Proposition 10

Suppose $h(z)$ is harmonic (or analytic) on $D(z_{0},R)$, and given by (2.40).
Then

(2.30)
$\frac{dh}{dz}=\sum_{n=1}^{\infty}n\beta_{n}(z-z_{0})^{n-1},\qquad\frac{dh}{d\bar{z}}=\sum_{n=1}^{\infty}n\beta_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n-1}.$

On the other hand, we are now in a position to prove

###### Proposition 11

If $h(a)=E_{a}[h(B_{\tau})]$ for any appropriate stopping time $\tau$, then
$h$ is harmonic.

Proof: The fact that $h(a)=E_{a}[h(B_{\tau})]$ means that the proof of
Poisson’s Integral formula (2.19) applies, and in particular $h$ can be
expressed in a series expansion of the form (2.40). It is now clear that
$\Delta h=4\frac{d^{2}h}{dzd\bar{z}}=0$

The reader will note that the condition that $h$ is analytic is relegated to
parentheses in the statements of Theorem 6 and Proposition 10; this is to
signify that, while these results do hold for analytic functions, in fact much
more is true. If $f$ is analytic at $z_{0}$, then the limit in (2.7) must
exist, regardless of the manner in which $\Delta z\longrightarrow 0$. In
particular, we may let $\Delta z=x$ be real, separate $f$ into real and
imaginary parts $u+iv$, and obtain

(2.31) $\begin{split}f^{\prime}(z_{0})&=\lim_{x\longrightarrow
0}\frac{f(z_{0}+x)-f(z_{0})}{x}=\lim_{x\longrightarrow
0}\frac{u(z_{0}+x)-u(z_{0})}{x}+\lim_{x\longrightarrow
0}\frac{v(z_{0}+x)-v(z_{0})}{x}\\\
&=\frac{du}{dx}(z_{0})+i\frac{dv}{dx}(z_{0}).\end{split}$

We may then conduct the similar operation where $\Delta z=iy$ is purely
imaginary, to obtain

(2.32) $\begin{split}f^{\prime}(z_{0})&=\lim_{y\longrightarrow
0}\frac{f(z_{0}+yi)-f(z_{0})}{yi}=\lim_{y\longrightarrow
0}\frac{u(z_{0}+yi)-u(z_{0})}{yi}+\lim_{y\longrightarrow
0}\frac{v(z_{0}+yi)-v(z_{0})}{yi}\\\
&=-i\frac{du}{dy}(z_{0})+\frac{dv}{dy}(z_{0}).\end{split}$

Equating real and imaginary parts of (2.31) and (2.32) gives the Cauchy-
Riemann equations:

(2.33) $\frac{du}{dx}=\frac{dv}{dy},\qquad\frac{du}{dy}=-\frac{dv}{dx}.$

Analytic functions satisfy these equations at all points in their domains.
Note that we may write

(2.34)
$\frac{df}{d\bar{z}}=\frac{du}{d\bar{z}}+i\frac{dv}{d\bar{z}}=\frac{1}{2}\Big{(}\frac{du}{dx}-\frac{dv}{dy}\Big{)}+\frac{i}{2}\Big{(}\frac{du}{dy}+\frac{dv}{dx}\Big{)}=0.$

Thus, the second series in (2.30) is identically 0 for analytic functions,
which means that all coefficients $\beta_{-n}$ must be zero as well. Theorem 6
and Proposition 10 are therefore eclipsed for analytic functions by

###### Theorem 7

Let $f(z)$ be analytic on $D(z_{0},R)$. Then we may write

(2.35) $f(z)=\sum_{n=0}^{\infty}\beta_{n}(z-z_{0})^{n}.$

This series converges uniformly on $D(z_{0},R^{\prime})$ for any
$R^{\prime}<R$. Furthermore, $f^{\prime}$ is analytic as well, and we have

(2.36)
$f^{\prime}(z)=\frac{df}{dz}=\sum_{n=1}^{\infty}n\beta_{n}(z-z_{0})^{n-1}.$

We also have $\frac{df}{d\bar{z}}=0$.

Remark: The relationship between harmonic and analytic functions can now be
clarified. If $f$ is analytic then the series expansion (2.35) holds, and thus
$Re(f)=\frac{f+\bar{f}}{2}$ and $Im(f)=\frac{f-\bar{f}}{2i}$ are both power
series in $z$ and $\bar{z}$ and are therefore harmonic. On the other hand,
if$f=u+iv$ is $C^{2}$ and satisfies the Cauchy-Riemann equations, then

(2.37)
$\frac{d^{2}u}{dx^{2}}+\frac{d^{2}u}{dy^{2}}=\frac{d^{2}v}{dydx}-\frac{d^{2}v}{dxdy}=0,\qquad\frac{d^{2}v}{dx^{2}}+\frac{d^{2}v}{dy^{2}}=-\frac{d^{2}u}{dydx}+\frac{d^{2}u}{dxdy}=0,$

so that $u$ and $v$ are each individually harmonic. $f$ is therefore harmonic
and admits the expansion given in (2.40), but with all coefficients
$\beta_{-n}$ equal to 0 by the Cauchy-Riemann equations; it is therefore
analytic.

###  2.5 Exercises

1 Show that Theorem 3 fails for the function $u(z)=u(x+yi)=y$ for the stopping
time $T_{K}$ where $K=\\{y=-1\\}$.

2 Let $T_{k}=\inf\\{t\geq 0:Im(B_{t})=k\\}$ and let $\hat{T}_{-1}=\sup\\{0\leq
t\leq T_{-2}:Im(B_{t})=-1\\}$. Let $\tau=T_{1}\wedge\hat{T}_{-1}$. Note that
$E[\tau]<\infty$ but $\tau$ is not a stopping time. Show that Dynkin’s formula
fails for $\tau$ with the function $u(x+yi)=y$.

3 Show that Möbius transformations are bijections from the Riemann sphere to
itself. Check also that they are analytic at $\infty$, and at $-d/c$ when
$c\neq 0$.

4 $SL_{2}({\mathbb{C}})$ is the space of all $2\times 2$ matrices with complex
entries and determinant 1. This is a group under matrix multiplication. The
Möbius group is the set of all Möbius transformations $\frac{az+b}{cz+d}$ with
$ad-bc=1$. This is a group under composition. Show that the map $\Theta$
defined by $\Theta(\frac{az+b}{cz+d})=\left(\begin{array}[]{ccc}a&b\\\
c&d\end{array}\right)$ is an isomorphism.

5 Let $\psi_{a}(z)=\frac{z-a}{1-\bar{a}z}$ with $|a|<1$. Show that
$|\psi_{a}(z)|\leq 1$, with equality precisely when $|z|=1$.

6 Prove the following equivalent forms for the Poisson kernel:

(2.38) ${\cal
P}_{r}(\theta)=\frac{1-r^{2}}{1-2r\cos\theta+r^{2}}=\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}=Re\Big{(}\frac{1+re^{i\theta}}{1-re^{i\theta}}\Big{)}$

7 Verify $(ii)-(v)$ of Lemma 2.

8 Prove Proposition 6 by employing the result of Exercise 1.4 from Chapter 1.

9

$a)$ Show that, if $r$ is an integer, then

(2.39)
$\int_{0}^{2\pi}e^{irt}dt=\left\\{\begin{array}[]{ll}2\pi&\qquad\mbox{if
}r=0\\\ 0&\qquad\mbox{if }r\neq 0\end{array}\right.$

$b)$ Suppose

(2.40)
$h(z)=\beta_{0}+\sum_{n=1}^{\infty}\beta_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}+\sum_{n=1}^{\infty}\beta_{n}(z-z_{0})^{n},$

for all $z\in D(z_{0},R)$. Use part (a) to show that

(2.41) $\beta_{n}=\frac{1}{2\pi
r^{n}}\int_{0}^{2\pi}h(z_{0}+re^{it})e^{-int}dt.$

Show also that $h$ is real-valued if, and only if,
$\beta_{-n}=\bar{\beta}_{n}$ for all $n$.

10 A function $f$ is anti-analytic if $\bar{f}$ is analytic. Show that if $f$
is a nonconstant, anti-analytic function, then $f(B_{t})$ is a time changed
Brownian motion, and identify the time change.

11 Suppose that ${\cal C}_{1},{\cal C}_{2}$ are two circles on the sphere
which do not intersect (so, they may be two circles in the plane or one line
and one circle). Show that there is a Möbius transformation taking them to two
concentric circles (i.e. circles with the same center) in the plane.
(Solution: Take one circle to the imaginary axis, and then translate and scale
so that the other has center on ${\mathbb{R}}$ and intersects ${\mathbb{R}}$
at $\frac{1}{r}$ and $r$. Then $\phi(z)=\frac{z-1}{z+1}$ does the trick.

##  Chapter 3 Examples and basic properties of harmonic and analytic functions

###  3.1 General properties of harmonic functions

We begin by studying
$h(z)=\ln|z|=\frac{1}{2}\ln|z|^{2}=\frac{1}{2}\ln(x^{2}+y^{2})$. A
straightforward calculation reveals $h$ to be harmonic (showing
$\frac{d^{2}h}{dx^{2}}+\frac{d^{2}h}{dy^{2}}=0$ is not difficult, but the
slick way is to calculate $\frac{dh}{dzd\bar{z}}$ and use Lemma 2). Thus, for
$r<R$ the function $h_{r}^{R}(z)=\frac{\ln R-\ln|z|}{\ln R-\ln r}$ is harmonic
on $A=A_{r,R}:=\\{r<|z|<R\\}$, and it is equal to 1 on $\\{|z|=r\\}$ and 0 on
$\\{|z|=R\\}$. Fix $a$ with $r<|a|<R$; by Itō’s formula we have
$h_{r}^{R}(a)=E_{a}[h_{r}^{R}(B_{T_{A}})]=P(|B_{T_{A}}|=r)$. We therefore
obtain the probability that $B_{t}$ exits $A$ at the inner circle
$\\{|z|=r\\}$ when started at $a$. We can now derive two important facts about
planar Brownian motion by this observation. The first to notice is that, upon
letting $R\nearrow\infty$, we see that $h_{r}^{R}(a)\nearrow 1$; this tells us
that $B_{t}$ is recurrent: regardless of the starting point, $B_{t}$ will hit
any open set eventually with probability $1$. On the other hand, if we hold
$R$ fixed and let $r\searrow 0$, then $h_{r}^{R}(a)\searrow 0$ as well; this
tells us that $P_{a}(B_{t}=0\mbox{ for some }t>0)=0$, and the same holds if 0
is replaced by any $b\neq 0$. This may seem counterintuitive, since after all
$P(\cup_{z\in{\mathbb{C}}}\\{B_{t}=z\mbox{ for some }t>0\\})=P(B_{t}\mbox{
hits some point in }{\mathbb{C}})=1$, but since there are uncountably many
points in ${\mathbb{C}}$ we do not run into a contradiction by summing
probabilities. This curious property of Brownian motion is commonly expressed
by a phrase like, ”planar Brownian motion is recurrent but doesn’t see
points”. We also can note that this phenomenon sheds new light the holomorphic
invariance of Brownian motion, Theorem 4, as the reader may have noticed a
potential problem with the intuitive proof given for that result: a holomophic
function is not a local dilation at a point at which its derivative is 0.
However, we will see that the zero set of a holomorphic function is a
countable set, and therefore Brownian motion avoids the zero set of the
derivative with probability 1, and again we do not run into trouble.

Let us calculate the power series for $h(z)$ around $z=1$. We have

(3.1) $\frac{d}{dz}\Big{(}\frac{1}{2}\ln
z\bar{z}\Big{)}=\frac{1}{2z}=\frac{1}{2(1+(z-1))}=\frac{1}{2}(1-(z-1)+(z-1)^{2}-\ldots).$

Proposition 10 tells us that integrating this series will give us the $z$-part
of the series for $h$, while the final statement of Theorem 6 and the fact
that $h$ is real valued tell us that the $\bar{z}$-part is obtained by
conjugation. Using $h(1)=0$, we arrive at the following expression:

(3.2)
$h(z)=\ln|z|=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-1)}}^{n}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(z-1)^{n}.$

The reader will note that this reduces to the familiar series $\ln
x=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^{n}$ when $z=x$ is real.

Let us now consider the function $\tilde{h}(z)=\mbox{Arg
}z=\tan^{-1}\frac{y}{x}=\tan^{-1}-i(\frac{z-\bar{z}}{z+\bar{z}})$, defined on
${\mathbb{C}}\backslash(-\infty,0]$. Another straightforward calculation shows
that $\Delta\tilde{h}=4\frac{d^{2}\tilde{h}}{dzd\bar{z}}=0$, and $\tilde{h}$
is harmonic. Suppose that $-\pi<\theta_{1}<\theta_{2}<\pi$, and suppose
$\theta_{1}<\mbox{Arg }a<\theta_{2}$. If we set
$\tilde{h}_{\theta_{1}}^{\theta_{2}}(z)=\frac{\theta_{2}-\mbox{Arg
}z}{\theta_{2}-\theta_{1}}$, we see that $\tilde{h}_{\theta_{1}}^{\theta_{2}}$
is equal to 1 on $\\{\mbox{Arg }z=\theta_{1}\\}$ and equal to 0 on
$\\{\mbox{Arg }z=\theta_{2}\\}$. Thus, if we set
$W=W_{\theta_{1}}^{\theta_{2}}=\\{\theta_{1}<\mbox{Arg }z<\theta_{2}\\}$, then
$\tilde{h}_{\theta_{1}}^{\theta_{2}}(a)=E_{a}[h_{\theta_{1}}^{\theta_{2}}(B_{T_{W}})]=P_{a}(\mbox{Arg
}(B_{T_{W}})=\theta_{1})$ (there is a subtlety here: since the domain $W$ is
unbounded, there are many harmonic functions with the given boundary values;
however, in Chapter LABEL:exitime we will see that there is only one such
function which remains bounded, and this must then be equal to
$P_{a}(\mbox{Arg }(B_{T_{W}})=\theta_{1})$). That is, we obtain the
probability that $B_{t}$ hits the set $\\{\mbox{Arg }z=\theta_{1}\\}$ before
hitting $\\{\mbox{Arg }z=\theta_{2}\\}$ when started at $a$, and this
probability is linear in $\mbox{Arg }a$.

Let us compute the power series for $\tilde{h}$ around $z=1$. We have

(3.3)
$\frac{d}{dz}\Big{(}\tan^{-1}i(\frac{z-\bar{z}}{z+\bar{z}})\Big{)}=\frac{-i}{2z}=\frac{-i}{2(1+(z-1))}=\frac{-i}{2}(1-(z-1)+(z-1)^{2}-\ldots).$

As before, Proposition 10 tells us that the $z$-part of the series for
$\tilde{g}$ comes by integrating this series, while the $\bar{z}$-part is
again obtained by conjugation. Again we have $\tilde{h}(1)=0$, so we obtain

(3.4) $\tilde{h}(z)=\mbox{Arg
}z=\frac{i}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-1)}}^{n}+\frac{-i}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(z-1)^{n}.$

This looks suspiciously like the series for $\ln|z|$, and we arrive at an
important idea. We note that we can write

(3.5) $h(z)+i\tilde{h}(z)=\ln|z|+i\mbox{Arg
}z=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(z-1)^{n},$

and Theorem 7 tells us that we have a holomorphic function. Thus, the
seemingly unrelated harmonic functions $h$ and $\tilde{h}$ have an intimate
relationship with each other, namely that they can be taken to be the real and
imaginary parts of a holomorphic function. We will call any such pair harmonic
conjugates. The reader may note a seeming incongruity with this example,
however, which is that the functions $h$ and $\tilde{h}$ do not have the same
domains of definition: $h$ is harmonic on ${\mathbb{C}}\backslash\\{0\\}$,
while $\tilde{h}$ can only be defined on ${\mathbb{C}}\backslash(-\infty.0]$,
as the function $\mbox{Arg }z$ cannot be made to be continuous on
$(-\infty.0]$.

We are led immediately to a fundamental question: suppose $h(z)$ is any real-
valued harmonic function; can we find another real-valued harmonic function
$\tilde{h}(z)$ such that $h+i\tilde{h}$ is analytic? We note that the previous
example shows that this is not always the case for all domains, however we can
show easily using power series that it is the case when the domain of
definition is a disk.

###### Theorem 8

Suppose $h(z)$ is a harmonic function on $D(z_{0},R)$. Then there is another
harmonic function $\tilde{h}(z)$ such that $h+i\tilde{h}$ is holomorphic on
$D(z_{0},R)$. This function is unique up to adding a constant.

Proof: Although the argument is implicit in the example of the logarithm, we
prove it in generality here. We know that we may write

(3.6)
$h(z)=\beta_{0}+\sum_{n=1}^{\infty}\beta_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}+\sum_{n=1}^{\infty}\beta_{n}(z-z_{0})^{n}.$

$\tilde{h}$ must now be chosen such that the power series for $h+i\tilde{h}$
contains no $\bar{z}$-part. If we write

(3.7)
$\tilde{h}(z)=\alpha_{0}+\sum_{n=1}^{\infty}\alpha_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}+\sum_{n=1}^{\infty}\alpha_{n}(z-z_{0})^{n},$

we see that we must have

(3.8)
$\sum_{n=1}^{\infty}\beta_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}+i\sum_{n=1}^{\infty}\alpha_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}=0,$

and we must therefore have $\alpha_{-n}=i\beta_{-n}$. Having obtained this,
the requirement that $\tilde{h}$ be real-valued forces
$\alpha_{n}=\bar{\alpha}_{-n}=-i\bar{\beta}_{-n}=-i\beta_{n}$. We therefore
obtain the unique (up to choice of $\alpha_{0}$) harmonic conjugate as

(3.9)
$\tilde{h}(z)=\alpha_{0}+i\sum_{n=1}^{\infty}\beta_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}-i\sum_{n=1}^{\infty}\beta_{n}(z-z_{0})^{n},$

and we have

(3.10)
$h+i\tilde{h}=\beta_{0}+\alpha_{0}+2\sum_{n=1}^{\infty}\beta_{n}(z-z_{0})^{n}.$

We will see later that a disk may be replaced by any simply connected domain.

Let us now establish some other properties of harmonic functions.

###### Theorem 9

[Mean value property] Suppose $h$ is harmonic on a domain $W$, and the set
$\\{|z-a|\leq r\\}\subseteq W$. Then

(3.11) $h(a)=\frac{1}{2\pi}\int_{0}^{2\pi}h(a+re^{i\theta})d\theta.$

Proof: By Dynkin’s formula, $h(a)=E_{a}[h(B_{T(D(a,r))}]$, and by the rotation
invariance of Brownian motion the distribution of $B_{T(D(a,r))}$ on
$\\{|z-a|=r\\}$ is uniform. The result follows.

In other words, $h(a)$ is equal to the average of its values on a circle in
$W$ centered at $a$. Note that this result can also be obtained from the
Poisson Integral Formula in Chapter 1.

######  Theorem 10 (Maximum/minimum principle)

Suppose $W$ is a bounded domain. If $h(z)$ is harmonic on $W$ and continuous
on $cl(W)$ then

$\inf_{w\in\delta W}h(w)\leq h(z)\leq\sup_{w\in\delta W}h(w)$

for all $z\in W$. Furthermore, if $h$ is nonconstant, then

$\inf_{w\in\delta W}h(w)<h(z)<\sup_{w\in\delta W}h(w)$

for all $z\in W$.

Note that if $W$ is the unit disk ${\mathbb{D}}$, then this result follows
from the Poisson Integral Formula (2.19) by virtue of the fact that
$\frac{1}{2\pi}\int_{0}^{2\pi}{\cal P}_{r}(t)dt=1$. The general result will be
shown in the next section.

######  Theorem 11 (Uniqueness principle)

Suppose $W$ is a bounded domain. If $h_{1}(z),h_{2}(z)$ are harmonic on $W$
and continuous on $cl(W)$ with $h_{1}(z)=h_{2}(z)$ for all $z\in\delta W$,
then $h_{1}(z)=h_{2}(z)$ for all $z\in W$.

Proof: Apply the maximum/minimum principle to the harmonic function
$h_{1}-h_{2}$.

###### Theorem 12

If $h(z)$ is harmonic and bounded on ${\mathbb{D}}^{\times}=\\{0<|z|<1\\}$,
then we can define $h(0)$ such that $h(z)$ so extended is harmonic on all of
${\mathbb{D}}$.

Proof: Dynkin’s formula implies that
$h(a)=E_{a}[h(B_{T({\mathbb{D}}^{\times})})]$. As was shown at the beginning
of this chapter, Brownian motion does not see points, and thus
$P_{a}(B_{T({\mathbb{D}}^{\times})}=0)=0$. The distribution of
$B_{T({\mathbb{D}}^{\times})}$ is therefore the same as that of
$B_{T({\mathbb{D}})}$, and thus $h(a)=E_{a}[h(B_{T({\mathbb{D}}^{\times})})]$.
As was shown in Chapter 1, $h(z)$ can therefore be expressed as a power series
in $z$ and $\bar{z}$, and this series provides the required extension to all
of ${\mathbb{D}}$.

###  3.2 The Dirichlet Problem

The Dirichlet problem in ${\mathbb{C}}$ is the following.

Dirichlet problem If $U$ is a bounded domain, and $h$ a continuous real-valued
function on $\delta U$, to extend $h$ to a harmonic function on $U$ which is
continuous on the closure of $U$.

As we will see, in many cases the Dirichlet problem may be solved by running a
Brownian motion. To be precise, we will show that $h(a)=E_{a}[h(B_{T(U)})]$
provides the desired solution whenever $U$ is a sufficiently nice domain.
However, there are domains, such as ${\mathbb{D}}^{\times}=\\{0<|z|<1\\}$, for
which the Dirichlet problem is not solvable. To see this, suppose that $h$ is
harmonic on ${\mathbb{D}}^{\times}$, continuous on its closure, with $h(z)=1$
for $|z|=1$ and $h(0)=0$. By Theorem 12, $h$ is in fact harmonic on all of
${\mathbb{D}}$, and thus by Theorem 11 $0=h(0)\geq\inf\\{|z|=1\\}h(z)=1$, a
contradiction. Nonetheless, we have the following classical result.

###### Theorem 13

The Dirichlet problem is solvable for any bounded simply connected domain $U$
in ${\mathbb{C}}$.

Proof: Since $U$ is bounded, $T(U)<\infty$ a.s. We can therefore extend $h$
from $\delta U$ to $cl(U)$ by setting $h(a)=E_{a}[h(B_{T(U)})]$, and we must
show that $h$ so extended on $U$ and continuous on $cl(U)$. Harmonicity
follows from the Strong Markov property:

$h(a)=E_{a}[h(B_{T(U)})]=E_{a}[E_{B_{T(D(a,r))}(\omega^{\prime})}[h(B_{T(U)}(\omega))]]=\frac{1}{2\pi}\int_{0}^{2\pi}h(a+re^{it})dt;$

note that in the final expectation in this string of equalities, the inner
expectation is taken with respect to $\omega\in\Omega$ while the outer one is
taken with respect to $\omega^{\prime}\in\Omega^{\prime}$, where the
probability spaces $\Omega$ and $\Omega^{\prime}$ are independent. To show
continuity we must prove that a Brownian motion started near a boundary point
of $U$ is likely to exit near that point, in particular

###### Lemma 3

If $y$ is a boundary point of a simply connected domain $U$, then given any
$\varepsilon>0$ we can find $\delta>0$ such that
$P_{a}(|B_{T(U)}-y|<\varepsilon)>1-\varepsilon$ whenever $|a-y|<\delta$.

The continuity of $h$ on $cl(U)$ now follows from from this by writing

$E_{a}[h(B_{T(U)})]=E_{a}[h(B_{T(U)})1_{\\{|B_{T(U)}-y|<\varepsilon\\}}]+E_{a}[h(B_{T(U)})1_{\\{|B_{T(U)}-y|\geq\varepsilon\\}}]$

and noting that for $a$ sufficiently close to $y$ the first term on the right
is as close as we like to $h(y)$ and the second can be made as small as we
like (using the boundedness of $h$).

Proof of Lemma 3: Consider the domain $V_{M}=\\{Re(z)<M\\}\backslash L$, where
$L$ is the ray in ${\mathbb{C}}$ emanating from $2\pi i$ horizontally to the
left; see Figure 3.1. We note that if $M$ is sufficiently large then
$P_{0}(B_{T(V_{M})}\in L)>1-\varepsilon$; this is evident, since as
$M\nearrow\infty$ we have $P_{0}(B_{T(V_{M})}\in L)\nearrow
P_{0}(T(V_{\infty})<\infty)$, where $V_{\infty}={\mathbb{C}}\backslash L$, and
$P_{0}(T(V_{\infty})<\infty)=1$ (proof: by Lévy’s Theorem, $arg(B_{t}-2\pi i)$
is a time-changed one-dimensional Brownian motion, is therefore recurrent, and
thus is equal to $-\pi$ at some finite time a.s.). We may therefore choose
$\delta$ sufficiently small so that $P_{0}(B_{T(V_{M})}\in L)>1-\varepsilon$
when $M=\ln\frac{\varepsilon}{2\delta}$, and ensure also that
$\delta<\frac{\varepsilon}{4}$. We claim now that $P_{a}(B_{T(U)}\in
B(y,\varepsilon))>1-\varepsilon$ whenever $|a-y|<\delta$.

Figure 3.1: The mapping by $f$ from $V_{M}$ to $\\{|z-y|<e^{M}\\}$.

To prove this, suppose that $|a-y|<\delta$, and let $w$ be the point in
$\delta U$ closest to $a$; note that $w$ is not necessarily the same as $y$,
but $|a-w|<\delta$ which implies by the triangle inequality that
$|y-w|<2\delta<\varepsilon$. Consider now the analytic function
$f(z)=(a-w)e^{z}+w$: it takes $0$ to $a$, takes $L$ to the line segment
$[a,w)$ connecting $a$ and $w$, and wraps the vertical line $\\{Re(z)=M\\}$
around the circle
$\\{|z-w|=|a-w|e^{M}=(\frac{|a-w|}{\delta})\frac{\varepsilon}{2}\\}$. We will
realize a Brownian motion starting at $a$ via Lévy’s Theorem as the image
under $f$ of a Brownian motion starting at $0$. The projection
$\tau=\sigma(T(V_{M}))$ under $f$ of the stopping time $T(V_{M})$ is again a
stopping time; we can describe it as ”the first time when either $B_{t}$ lies
on the line segment $(a,w)$ having wound once anticlockwise about $w$ or
$|B_{t}-w|=(\frac{|a-w|}{\delta})\frac{\varepsilon}{2}$”. Note that
$P_{a}(B_{\tau}\in(a,w))=P_{0}(B_{T(V_{M})}\in L)>1-\varepsilon$. Furthermore,
on the event $\\{B_{\tau}\in(a,w)\\}$ we may connect $B_{\tau}$ to $a$ by a
straight line segment and form a closed path which winds about the boundary
point $w$; if this path remained entirely within $U$ this would contradict
simple connectivity, and since $(a,w)$ lies within $U$ we conclude that
$B_{t}$ intersects $\delta U$ before $\tau$. Thus, on the event
$\\{B_{\tau}\in(a,w)\\}$ we have $T(U)<\tau\leq\inf\\{t\geq
0:|B_{t}-w|=(\frac{|a-w|}{\delta})\frac{\varepsilon}{2}\\}$. The triangle
inequality implies easily that
$\\{|z-w|<(\frac{|a-w|}{\delta})\frac{\varepsilon}{2}\\}\subseteq\\{|z-w|<\frac{\varepsilon}{2}\\}\subseteq\\{|z-y|<\varepsilon\\}$,
and we conclude that $P_{a}(T(U)<\inf\\{t\geq
0:|B_{t}-y|=\varepsilon\\})>1-\varepsilon$. Since $\\{T(U)<\inf\\{t\geq
0:|B_{t}-y|=\varepsilon\\}\\}\subseteq\\{|B_{T(U)}-y|<\varepsilon\\}$, the
lemma is proved.

Remarks:

  * •

The ray $L$ can be replaced in the proof by the union of the rays emanating to
the left from all points of the form $n2\pi i$ for integer $n\neq 0$. This
would form an even smaller stopping time $\tau$, but would not change the
proof in any meaningful way.

  * •

It may be interesting to note that the choice of $\delta$ depends only on
$\varepsilon$ and not on $U$.

  * •

The proof given here can be easily extended to other situations, such as when
the complement of $U$ is formed of finitely many connected components, none of
which is a single point.

  * •

The boundedness requirement on $U$ can in many cases be relaxed, but the
Dirichlet problem and even the maximum/minimum and uniqueness principles for
harmonic functions are rather subtle for unbounded domains. See Chapter 8 for
more details on this.

###  3.3 The Identity Theorem, Open Mapping Theorem, and Maximum Principle

We now discuss a number of results on general analytic and harmonic function.
We begin with a result, commonly referred to as the Identity Theorem, which
states that if two analytic functions coincide on a set which contains an
accumulation point than they agree everywhere. Note that this is not true for
real-valued differentiable functions on the line, as for instance the function
$f(x)=e^{-1/x}$ for $x>0$ and $f(x)=0$ for $x\leq 0$ is $C^{\infty}$ on the
real line, agrees with the constant function 0 on the negative real axis, and
yet is clearly not equal to 0 for all $x$.

###### Theorem 14

Let $f$ be a nonconstant analytic function on a domain $W$, and let ${\cal Z}$
be the zero set of $f$; that is, let ${\cal Z}=\\{z\in W:f(z)=0\\}$. Then
${\cal Z}$ is a discrete set in $W$.

Proof: Let $z_{0}\in{\cal Z}$. We know that in some disk centered at $z_{0}$
we can write $f(z)=\sum_{n=1}^{\infty}\beta_{n}(z-z_{0})^{n}$. Let $N\geq 1$
be the smallest index so that $\beta_{N}\neq 0$; we can then write
$f(z)=(z-z_{0})^{N}\sum_{n=N}^{\infty}\beta_{n}(z-z_{0})^{n-N}$. As
$z\longrightarrow z_{0}$,
$\sum_{n=N}^{\infty}\beta_{n}(z-z_{0})^{n-N}\longrightarrow\beta_{N}\neq 0$,
and it follows by continuity that this expression is nonzero on some (possibly
smaller) disk centered at $z_{0}$. Since $(z-z_{0})^{N}$ is only 0 at $z_{0}$,
we see that $f$ is not zero in this disk except at $z_{0}$, and $z_{0}$ is
therefore not an accumulation point of ${\cal Z}$.

###### Corollary 4

If $f,g$ are analytic on a domain $W$, and the set $\\{z:f(z)=g(z)\\}$ has an
accumulation point in $W$, then $f(z)=g(z)$ for all $z$ in $W$.

Proof: Apply Theorem 14 to the analytic function $f-g$.

We now will prove that the image under a nonconstant analytic function of an
open set is again open. This is commonly known as the Open Mapping Theorem.
Note that, again, nothing like this is true for real-valued differentiable
functions on the line, as for instance the image of the real line under the
sine function is the closed interval $[-1,1]$. First, we need a lemma about
Brownian motion.

###### Lemma 4

Let $a\in{\mathbb{C}}$, and for any $\delta>0$ let
$\tau_{\delta}=\inf\Big{\\{}t\geq 0:B_{t}\in\\{|z-a|=\delta\\}\Big{\\}}$. Fix
$r>0$, and let $V$ be an open set contained in $D(a,r)$, where
$D(a,r)=\\{|z-a|<r\\}$. Let $B_{t}$ be a Brownian motion starting at $a$. Then
$P(B_{t}\in V\mbox{ for some }t\leq\tau_{r})>0$.

Proof: We have

(3.12) $P_{a}(B_{t}\in V\mbox{ for some }t\leq\tau_{r})\geq
P_{a}(B_{t}\in{\cal C}\mbox{ for some }t\leq\tau_{r})\geq
P_{a}(B_{\tau_{r^{\prime}}}\in{\cal C})=\frac{\theta_{2}-\theta_{1}}{2\pi}.$

###### Theorem 15

Let $f$ be a nonconstant analytic function on a domain $W$. Then $f(W)$ is a
domain.

Proof: We must prove that $f(W)$ is open and connected. Analyticity is not
needed for connectedness: any continuous image of a connected set is
connected, since if $A$ and $B$ disconnect $f(W)$ then $f^{-1}(A)$ and
$f^{-1}(B)$ disconnect $W$. Openness is a bit more tricky. Let $v\in f(W)$,
and choose $a\in W$ such that $f(a)=v$. By Corollary 4 we can choose a small
$r>0$ such that $\bar{D}(a,r)\subseteq W$ and $f(w)\neq v$ on $\\{|w-a|=r\\}$.
If we start a Brownian motion at $a$ and stop it at the hitting time $\tau$ of
$\\{|w-a|=r\\}$, then $\hat{B}_{t}=f(B_{C_{t}})$ is a Brownian motion starting
at $v$ and stopped at the stopping time $C^{-1}_{\tau}$, which satisfies
$f(\hat{B}_{C^{-1}_{\tau}})=f(B_{\tau})\in f(\\{|w-a|=r\\})$. Let $\gamma$
denote the curve $f(\\{|w-a|=r\\})$. In traveling from $v$ to $\gamma$,
$\hat{B}_{t}$ must cross first the circle $\\{|w-v|=m\\}$, where
$m=\inf\\{|w-v|:w\in\gamma\\}>0$; so if we let $\hat{\tau}$ be the first time
$\hat{B}_{t}$ hits $\\{|w-v|=m\\}$ then $C^{-1}_{\tau}\geq\hat{\tau}$ a.s.

Lemma 4 now shows that $\hat{B}_{t}=f(B_{C_{t}})$ hits any open set in
$\\{|w-v|<m\\}$ with positive probability before time $\hat{\tau}$, and this
shows in particular that $\\{|w-v|<m\\}$ is contained in the closure of
$f(\bar{D}(z,r))$. However, $\bar{D}(z,r)$ is a compact set, so its image
under the continuous map $f$ is compact and thus closed, and therefore
$\\{|w-v|<m\\}\subseteq f(\bar{D}(z,r))\subseteq f(W)$. This proves that
$f(W)$ is open.

###### Theorem 16

Let $f$ be a nonconstant analytic function on a domain $W$. Then $f$ does not
attain its maximum value on $W$; that is, there is no point $z$ such that
$|f(z)|=\sup_{w\in W}|f(w)|$ for all $w\in W$.

Proof: Let ${\cal M}=\\{z\in W:|f(z)|=\sup_{w\in W}|f(w)|\\}$. ${\cal M}$ is
closed, being the preimage of a point under the continous map $f$; but we may
argue that it is open as well, as follows. For $a\in{\cal M}$, choose a small
$r>0$ such that $\bar{D}(a,r)\subseteq W$, and let $\tau$ be the first time
Brownian motion hits $\\{|w-a|=r\\}$. Then $|f(a)|=|E_{a}[f(B_{\tau})]|\leq
E_{a}[|f(B_{\tau})|]\leq\sup_{|w-a|=r}|f(w)|\leq|f(a)|$. All inequalities in
this string must then be equalities, which can only happen if $|f(w)|=|f(a)|$
almost surely on $\\{|w-a|=r\\}$. By continuity, $|f(w)|=|f(a)|$ for all $w$
on this circle, and by repeating this argument for any radii less than $r$ we
see that $|f(w)|=|f(a)|$ for all $w\in D(a,r)$. This proves ${\cal M}$ is an
open set, and since it is both open and closed the connectedness of $W$ now
proves that ${\cal M}=\emptyset$ or ${\cal M}=W$. If ${\cal M}=W$ then
$f(W)\subseteq\\{|w|=const.\\}$, but this implies by Theorem 15 that $f$ is
constant, a contradiction. Thus, ${\cal M}=\emptyset$, proving the theorem.

###### Corollary 5

Let $f$ be a nonconstant analytic function on a domain $W$ with $f(z)\neq 0$
on $W$. Then $f$ does not attain its minimum value on $W$; that is, there is
no point $z$ such that $|f(z)|=\inf_{w\in W}|f(w)|$ for all $w\in W$.

Proof: Apply Theorem 16 to the analytic function $1/f$.

We should mention that the Maximum Principle applies to real-valued harmonic
functions as well. The proof given here carries over directly, with the only
difference being that dealing with the modulus is no longer necessary, and
this removes the requirement that the function never be zero to deduce the
Minimum Principle.

###### Theorem 17

Let $h$ be a nonconstant harmonic function on a domain $W$. Then $f$ does not
attain its maximum (resp. minimum) value on $W$; that is, there is no point
$z$ such that $h(z)=\sup_{w\in W}h(w)$ (resp. $h(z)=\inf_{w\in W}h(w)$) for
all $w\in W$.

###  3.4 The exponential and trigonometric functions

Let us examine the exponential function $e^{z}=e^{x+yi}=e^{x}(\cos y+i\sin y)$
more closely. This is an analytic function mapping the plane to itself, with
$|e^{z}|=e^{x}$ and $\mbox{Arg }(e^{z})=y$. We see that this function and the
logarithm defined earlier in the chapter are inverses wherever Log  can be
applied, in the sense that $e^{\mbox{Log }z}=e^{\ln|z|+i\mbox{Arg
}z}=|z|e^{i\mbox{Arg }z}=z$, and $\mbox{Log }e^{z}=\ln e^{x}+i\mbox{Arg }(\cos
y+i\sin y)=x+\tilde{y}i$, where $\tilde{y}$ is the value in $(-\pi,\pi]$ such
that $y=\tilde{y}(\mbox{mod }2\pi)$. Since the argument of $e^{z}$ can be any
value, and the modulus of $e^{z}$ can be any positive value, the image of the
plane under this map is ${\mathbb{C}}\backslash\\{0\\}$. The vertical line
$\\{y=y_{0}\\}$ in the plane gets mapped to a curve of constant argument; that
is, to the infinite ray connecting 0 to $\infty$, which can be parameterized
as $\\{te^{iy_{0}}:0<t<\infty\\}$. The horizontal line $\\{x=x_{0}\\}$ in the
plane gets mapped to a curve of constant modulus; that is, to the circle
centered at 0 of radius $e^{x_{0}}$. This geometric understanding of the
function gives rise to several useful consequences for Brownian motion, as we
now describe.

To begin with, it is a standard fact that one-dimensional Brownian motion is
recurrent; in fact we showed in Section 3.1 that planar Brownian motion is
recurrent, and it is immediate from this that the one-dimensional variety is
as well by considering the real or imaginary part of the planar case. However,
there are proofs of recurrence in one dimension which do not use work done in
the plane. We may therefore deduce planar recurrence by noting that recurrence
on the line shows that the real part of a Brownian motion $B_{t}$ starting at
any point $x_{0}+iy_{0}$ will eventually hit any other point $x_{1}$, which
means $B_{t}$ hits $\\{x=x_{1}\\}$ with probability 1. Taking the image under
the exponential map, $B_{t}$ is mapped to a time changed Brownian motion
$\tilde{B}_{t}$, and $\\{x=x_{1}\\}$ is mapped to $\\{|z|=e^{x_{1}}\\}$. We
conclude that

(3.13) $P_{e^{x_{0}+y_{0}i}}(\tilde{B}_{t}\in\\{|z|=e^{x_{1}}\\}\mbox{ for
some }t>0)=1,$

which is merely a restatement of recurrence. However, with a bit more work we
can obtain an extension of this result which will be important to us. Let
$r<R$ be two positive values, and define sequences of stopping times
$\tau^{(r)}_{n}$ and $\tau^{(R)}_{n}$ as follows. Let
$\tau^{(r)}_{1}=\inf\\{t>0:B_{t}\in\\{|z|=r\\}\\}$, then
$\tau^{(R)}_{1}=\inf\\{t>\tau^{(r)}_{1}:B_{t}\in\\{|z|=R\\}\\}$. For larger
$n$ define these similarly as
$\tau^{(r)}_{n}=\inf\\{t>\tau^{(R)}_{n-1}:B_{t}\in\\{|z|=r\\}\\}$ and
$\tau^{(R)}_{n}=\inf\\{t>\tau^{(r)}_{n}:B_{t}\in\\{|z|=R\\}\\}$. In words,
$\tau^{(r)}_{1}$ is the first time $B_{t}$ hits $\\{|z|=r\\}$, then
$\tau^{(R)}_{1}$ is the first time after $\tau^{(r)}_{1}$ that $B_{t}$ hits
$\\{|z|=R\\}$, then $\tau^{(r)}_{1}$ is the first time after $\tau^{(R)}_{1}$
that $B_{t}$ hits $\\{|z|=r\\}$, etc. Recall Theorem 1, the Strong Markov
Property, from Chapter 1, which states that a Brownian motion behaves
identically following a stopping time to a Brownian motion starting at time 0;
together with the recurrence already established this guarantees that the
random variables $\tau^{(r)}_{1},\tau^{(R)}_{n}-\tau^{(r)}_{n},$ and
$\tau^{(r)}_{n}-\tau^{(R)}_{n-1}$ are all finite almost surely, regardless of
the starting point of $B_{t}$. This in turn guarantees the almost sure
finiteness of $\tau^{(r)}_{n}$ and $\tau^{(R)}_{n}$ for all n; that is,
$B_{t}$ will make infinitely many trips back and forth between $\\{|z|=r\\}$
and $\\{|z|=R\\}$. Let a small $\delta>0$ be given, and apply our knowledge
with $r=\delta/2,R=\delta$. Note that for the duration of a trip between
$\\{|z|=\delta/2\\}$ and $\\{|z|=\delta\\}$ the Brownian motion $B_{t}$ must
reside entirely within the disk $D(0,\delta)$. Thus, we have

(3.14)
$\int_{0}^{\infty}1_{D(0,\delta)}(B_{s})ds\geq\sum_{n=1}^{\infty}(\tau^{(\delta)}_{n}-\tau^{(\delta/2)}_{n}).$

Set $Y_{n}=(\tau^{(\delta)}_{n}-\tau^{(\delta/2)}_{n})$. It is not hard to see
by the Strong Markov Property and rotation invariance that the $Y_{n}$’s form
a sequence of independent, identically distributed, positive random variables.
We therefore have, for any $a\in{\mathbb{C}}$,

(3.15) $E_{a}[e^{-\int_{0}^{\infty}1_{D(0,\delta)}(B_{s})ds}]\leq
E_{a}[e^{-\sum_{n=1}^{\infty}Y_{n}}]=\lim_{n\longrightarrow\infty}E_{a}[e^{-Y_{1}}]^{n}=0.$

This can only occur if $\int_{0}^{\infty}1_{D(0,\delta)}(B_{s})ds=\infty$ a.s.
By translating if necessary any open set $A$ so that it contains $0$, we have
proved the following.

###### Proposition 12

For Brownian motion starting at any point and for any open set $A$ in
${\mathbb{C}}$, we have

(3.16) $\int_{0}^{\infty}1_{A}(B_{s})ds=\infty\mbox{ a.s.}$

That is, Brownian motion spends an infinite amount of time in any open set
with probability 1. This will be important in the next chapter, when we study
general functions analytic on all of ${\mathbb{C}}$.

Now do winding and Spitzer.

###  3.5 Entire and almost entire functions

An entire function is a function analytic on all of ${\mathbb{C}}$. We will
call a function almost entire if it is analytic on a domain $W$ such that
$W^{c}$ is a countable, closed set. This is a natural definition in relation
to Brownian motion, since a Brownian motion started in $W$ will never leave
$W$, just as a Brownian motion can never leave the domain of an entire
function. If $f$ is entire or almost entire, then $f(B_{t})$ is a time-changed
Brownian motion, which means that its paths are precisely the paths of a
Brownian motion, provided that we can show that nothing strange happens with
the time change $\sigma_{t}=\int_{0}^{t}|f^{\prime}(B_{s})|^{2}ds$; that is,
we need to know that $\sigma_{t}$ doesn’t blow up for some $t$, and also
doesn’t approach some finite limit as $t\longrightarrow\infty$ (which would
correspond to the image Brownian motion $f(B_{t})$ stopping). That is, we need
the following result.

###### Theorem 18

Let $f$ be almost entire and nonconstant. Then
$\sigma_{t}=\int_{0}^{t}|f^{\prime}(B_{s})|^{2}ds$ is almost surely finite for
any $t$, and $\sigma_{t}\nearrow\infty$ almost surely as
$t\longrightarrow\infty$.

Proof: Since $f$ is nonconstant, $f^{\prime}$ is not identically 0, which
means we can choose $\varepsilon>0$ so that the open set
$A=\\{z:|f^{\prime}(z)|>\varepsilon\\}$ is nonempty. Then

(3.17)
$\int_{0}^{\infty}|f^{\prime}(B_{s})|^{2}ds\geq\int_{0}^{\infty}\varepsilon^{2}1_{A}(B_{s})ds=\varepsilon^{2}\int_{0}^{\infty}1_{A}(B_{s})ds=\infty\mbox{
a.s.}$

by Proposition 12. This proves the second part of the statement, and the first
is immediate, since $f^{\prime}$ is continuous and thus bounded on the a.s.
compact sets $\\{B_{s}:0\leq s\leq t\\}$.

We can use this fact to obtain some results on entire and almost entire
functions. For instance, since Brownian motion is recurrent and the paths of
$f(B_{t})$ are the paths of a Brownian motion, $f(B_{t})$ must be recurrent as
well. This gives

###### Theorem 19

Let $f$ be nonconstant and entire or almost entire. Then the image of $f$ is
dense in ${\mathbb{C}}$.

This gives several immediate corollaries which are quite important.

######  Corollary 6 (Liouville’s Theorem)

A bounded entire or almost entire function is a constant.

######  Corollary 7 (Fundamental Theorem of Algebra)

Suppose $p(z)=\beta_{k}z^{k}+\ldots+\beta_{0}$ is a degree $k$ polynomial,
with $\beta_{0},\ldots,\beta_{k}\in{\mathbb{C}}$ and $\beta_{k}\neq 0$. Then
$p$ has a root; that is, there is an $a\in{\mathbb{C}}$ such that $f(a)=0$.

Proof: Since $p({\mathbb{C}})$ is dense in ${\mathbb{C}}$, we can find a
sequence of points $\\{z_{n}\\}$ such that $p(z_{n})\longrightarrow 0$. The
compactness of $\hat{\mathbb{C}}$ allows us to extract a convergent
subsequence, and passing to this subsequence if necessary allows us to assume
that $z_{n}\longrightarrow a\in\hat{C}$. We must show that $a\neq\infty$. Note
that we can write

(3.18)
$|p(z)|=|z|^{k}\Big{|}\beta_{k}+\frac{\beta_{k-1}}{z}+\ldots+\frac{\beta_{0}}{z^{k}}\Big{|}.$

As $z\longrightarrow\infty$, we have
$|\beta_{k}+\frac{\beta_{k-1}}{z}+\ldots+\frac{\beta_{0}}{z^{k}}|\longrightarrow|\beta_{k}|\neq
0$, so that $|p(z)|\longrightarrow\infty$; thus, $a\neq\infty$. We then have
$a\in{\mathbb{C}}$, and by continuity we obtain
$f(a)=\lim_{n\longrightarrow\infty}f(z_{n})=0$.

In fact, $p(z)$ will have $k$ roots; see Exercise 3.6 below.

Theorem 19 is just the beginning for entire functions. We will show later
using Brownian motion that, if $f$ is entire, then $f({\mathbb{C}})$ is all of
${\mathbb{C}}$ with at most one point removed. This is known as Picard’s
Little Theorem.

###  3.6 Exercises

1 Suppose $p(z)=\beta_{k}z^{k}+\ldots+\beta_{0}$ is a degree $k$ polynomial,
with $\beta_{0},\ldots,\beta_{k}\in{\mathbb{C}}$ and $\beta_{k}\neq 0$. Show
that $p$ has $k$ roots, counting multiplicities. That is, show that there are
values $a_{1},\ldots,a_{k},c\in{\mathbb{C}}$ such that
$p(z)=c(z-a_{1})(z-a_{2})\ldots(z-a_{k})$. (Hint: If $p(a)=0$, consider
$p(z)-p(a)$).

##  Chapter 4 Simply connected domains

###  4.1 Contour integration and Cauchy’s Theorem

One of the most important theorems in complex analysis is Cauchy’s Theorem,
which is the subject of this section. Ordinarily, Cauchy’s Theorem is
formulated using contour integration, but there is an equivalent form which
will be more useful for our purposes. First, we need a definition.

A domain $W$ is simply connected if $W^{c}$ is connected, where the complement
is taken in $\hat{\mathbb{C}}$. It is important for this definition that
$\infty$ be considered a point in $W^{c}$, for the first domain in the picture
below may appear to have a disconnected complement, but in fact it is
connected in the sphere. Armed with this definition, we may state the theorem.

######  Theorem 20 (Cauchy’s Theorem)

Let $f$ be an analytic function on a simply connected domain $W$. Then there
is an analytic antiderivative of $f$ on $W$; that is, there is an analytic
function $F$ on $W$ such that $F^{\prime}(z)=f(z)$.

In order to prove this theorem, we introduce contour integration. Let
$\gamma(t)$ be a continuous, piecewise smooth function from an interval
$[a,b]\subseteq{\mathbb{R}}$ into ${\mathbb{C}}$ which is only non-smooth at
finitely many points in $[a,b]$; we will refer to such a function and its
image as a rectifiable curve. The image $\gamma([a,b])$ will be a connected
curve in ${\mathbb{C}}$, and we will abuse notation somewhat by simply
referring to this curve as $\gamma$. The contour integral along $\gamma$ (also
known as the complex line integral), denoted $\int_{\gamma}fdz$, is defined by
the formula

(4.1) $\int_{\gamma}f(z)dz=\int_{a}^{b}f(\gamma(t))\gamma^{\prime}(t)dt.$

It can be shown easily that this quantity does not depend on the
parameterization chosen (see Exercise 7.4). A curve $\gamma$ is closed if
$\gamma(a)=\gamma(b)$. The following is an equivalent form of Cauchy’s
Theorem.

######  Theorem 21 (Cauchy’s Theorem - contour integral form)

If $f(z)$ is analytic on a simply connected domain $W$, and $\gamma$ is a
closed path in $W$, then

(4.2) $\int_{\gamma}f(z)\,dz=0.$

The two forms of Cauchy’s Theorem are equivalent by the following proposition.

###### Proposition 13

A function $f$ on a domain $W$ has an analytic antiderivative on $W$ if, and
only if, $\int_{\gamma}f(z)\,dz=0$ for any closed curve $\gamma\subseteq W$.

Remark: Note that $W$ is not assumed to be simply connected here.

Proof of Proposition 13: Suppose that $f$ has an analytic antiderivative $F$
on $W$. Then, for any closed curve $\gamma$, we have

(4.3)
$\int_{\gamma}f(z)\,dz=\int_{a}^{b}f(\gamma(t))\gamma^{\prime}(t)dt=\int_{a}^{b}\Big{(}\frac{d}{dt}F(\gamma(t))\Big{)}dt=F(\gamma(b))-F(\gamma(a))=0.$

On the other hand, if (4.2) holds, then we may fix $z_{0}\in W$ and, for any
$z\in W$, define

(4.4) $F(z)=\int_{z_{0}}^{z}f(w)dw,$

where the notation $\int_{z_{0}}^{z}$ indicates that the integral is taken
over any curve traveling from $z_{0}$ to $z$. This is well defined, for if
$\gamma_{1},\gamma_{2}$ are two such such curves, then we can form a new curve
$\gamma$ which follows $\gamma_{1}$ from $z_{0}$ to $z$ and then follows
$\gamma_{2}$ backwards from $z_{0}$ to $z$; $\gamma$ is a closed curve, which
means that $\int_{\gamma}f(z)dz=0$, and the integral over $\gamma_{2}$
traversed backwards is equal to the integral over $\gamma_{2}$ traversed
forward (see Exercise 7.4); it follows that
$\int_{\gamma_{1}}f(z)dz=\int_{\gamma_{2}}f(z)dz$. With this notation, it is
clear from (4.2) and Exercise 7.4 that

(4.5)
$\int_{z_{0}}^{z_{1}}f(w)dw+\int_{z_{1}}^{z_{2}}f(w)dw+\int_{z_{2}}^{z_{0}}f(w)dw=0,$

since a curve traveling from $z_{0}$ to $z_{1}$ to $z_{2}$ and back to $z_{0}$
is closed, and thus

(4.6) $\begin{split}\lim_{h\longrightarrow
0}\frac{F(z+h)-F(z)}{h}&=\lim_{h\longrightarrow
0}\frac{\int_{z_{0}}^{z+h}f(w)dw-\int_{z_{0}}^{z}f(w)dw}{h}\\\
&=\lim_{h\longrightarrow 0}\frac{1}{h}\int_{z}^{z+h}f(w)dw.\end{split}$

For $h$ sufficiently small we can take this interval over the line segment
connecting $z$ and $z+h$, which we can parameterize as
$\gamma(t)=\\{z+th:0\leq t\leq 1\\}$. We then have

(4.7) $\lim_{h\longrightarrow 0}\frac{F(z+h)-F(z)}{h}=\lim_{h\longrightarrow
0}\frac{1}{h}\int_{0}^{1}f(z+ht)hdt=\lim_{h\longrightarrow
0}\int_{0}^{1}f(z+ht)dt=f(z).$

Thus, $F$ is analytic in $W$, and $F^{\prime}(z)=f(z)$.

######  Corollary 8 (Morera’s Theorem)

If a continuous function $f$ on a domain $W$ satisfies
$\int_{\gamma}f(z)\,dz=0$ for any closed curve $\gamma\subseteq W$ then $f$ is
analytic.

Proof: Proposition 13 shows that $f=F^{\prime}$ for an analytic function $F$,
and Theorem 7 then shows that $f$ is itself analytic.

We will now prove Cauchy’s Theorem; in particular, we will prove the integral
form, Theorem 21.

Proof of Theorem 21: We will progress through a sequence of cases, ending with
the general case.

Case 1: $\gamma$ is contained in a disk $D(z_{0},r)$ which is contained in
$W$. Here we expand $f$ in a power series around $z_{0}$,
$f(z)=\sum_{n=0}^{\infty}\beta_{n}(z-z_{0})^{n}$, and we are assured that the
radius of convergence of the series is at least $r$. If we let
$F(z)=\sum_{n=0}^{\infty}\frac{\beta_{n}}{n+1}(z-z_{0})^{n+1}$, then the
radius of convergence of the series for $F$ is the same as the series for $f$
(see Exercise 7.4), and we can differentiate termwise to get
$F^{\prime}(z)=f(z)$. By Proposition 13, $\int_{\gamma}f(z)dz=0$.

Case 2: $\gamma$ is a triangle. Exercise 7.4 shows that we can assume that
$\gamma$ is traversed counterclockwise. Here Case 1 does not necessarily
directly apply, but the idea is to express $\gamma$ as the sum of a finite
number of smaller triangles, each of which is contained in a disk contained in
$W$; note that it will be here that we make use of simple connectivity, since
it guarantees us that the interior of $\gamma$ is contained in $W$. Let
$\gamma_{1}^{1}=\gamma$, then let $\gamma_{1}^{2},\ldots,\gamma_{4}^{2}$ be
the four triangles obtained by connecting the midpoints of the sides of
$\gamma_{1}^{1}$, and then $\gamma_{1}^{3},\ldots,\gamma_{16}^{3}$ be the
sixteen triangles obtained by connecting the midpoints of the sides of each
$\gamma_{j}^{2}$, and so forth.

If we sum the integrals of $f$ over each of the triangles at a given level
traversed counterclockwise, we see that all but the outside line segments are
integrated over in both directions. Exercise 7.4 (ii) then shows that the
contribution of each of these interior segments vanishes, and we obtain

(4.8) $\int_{\gamma}f(z)dz=\sum_{j=1}^{4^{n-1}}\int_{\gamma_{j}^{n}}f(z)dz.$

Let $d(z,\delta W)=\inf\\{|z-w|:w\in\delta W\\}$. By compactness (see Exercise
7.4), there is an $\varepsilon>0$ such that $D(z,\varepsilon)\subseteq W$ for
all $z$ on $\gamma$ and in the interior of $\gamma$. We now need only choose
$n$ large enough so that the diameter of $\gamma_{j}^{n}$ is less than
$\varepsilon$, and we see that each of the triangles $\gamma_{j}^{n}$ is
contained in a disk contained in $W$. It follows from Case 1 that the integral
over each $\gamma_{j}^{n}$ is $0$, and thus $\int_{\gamma}f(z)dz=0$.

Case 3: $\gamma$ is a simple polygon; that is, $\gamma$ is a piecewise linear
curve with no self-intersections. Again we can assume that $\gamma$ is
traversed counterclockwise. It is not hard to show that any such polygon
admits a triangulation ${\cal T}$ (see Appendix Jordan Curve), and the
integral over each triangle is 0 by Case 2. If we sum the integrals over the
triangles traversed counterclockwise, then as in Case 2 each interior line
segment will be traversed once in each direction, and by canceling these
quantities we see that $\int_{\gamma}f(z)dz$ is equal to the sum of the
integrals over the triangles in ${\cal T}$. We conclude that
$\int_{\gamma}f(z)dz=0$.

Case 4: $\gamma$ is a polygonal path. Here $\gamma$ has a finite set of
intersections, and we can break $\gamma$ into the union of a finite number of
simple polygons. The integral over each of these components of $\gamma$ is 0,
and thus $\int_{\gamma}f(z)dz=0$.

Case 5: $\gamma$ is arbitrary. Again by compactness there is an
$\varepsilon>0$ such that $D(z,\varepsilon)\subseteq W$ for all $z$ on
$\gamma$, and we can choose $a=t_{0}<t_{1}<\ldots<t_{N}=b$ such that
$\gamma\subseteq\cup_{k=1}^{N}D(\gamma(t_{k}),\varepsilon)$. We can then
choose $z_{k}\in D(\gamma(t_{k}),\varepsilon)\cap
D(\gamma(t_{k+1}),\varepsilon)$ for $k=1,\ldots,N$. We now let
$\tilde{\gamma}$ be the polygonal path formed by taking the union of the line
segments from $z_{k}$ to $z_{k+1}$.

Since the line segment connecting $z_{k}$ and $z_{k+1}$ lies in the disk
$D(\gamma(t_{k+1}),\varepsilon)$, the integral over this segment is the same
as the integral over the portion of $\gamma$ connecting $z_{k}$ to $z_{k+1}$
by Case 1. It follows that $\int_{\gamma}f(z)dz=\int_{\tilde{\gamma}}f(z)dz$,
and since $\tilde{\gamma}$ is a polygonal path we have $\int_{\gamma}f(z)dz=0$
by Case 4. Thus, $\int_{\gamma}f(z)dz=0$. This completes the proof of Cauchy’s
Theorem.

We now deduce some important consequences.

###### Corollary 9

Let $f$ be an analytic function on a simply connected domain $W$, with
$f(z)\neq 0$ for all $z\in W$. Then there is an analytic logarithm of $f$ on
$W$; that is, there is an analytic function $g$ on $W$ such that $e^{g}=f$.

Proof: If $f\neq 0$ on $W$, Cauchy’s theorem tells us that we may find an
antiderivative $\tilde{g}$ of the analytic function $\frac{f^{\prime}}{f}$. We
then have

(4.9)
$\frac{d}{dz}fe^{-\tilde{g}}=f^{\prime}e^{-\tilde{g}}+fe^{-\tilde{g}}\tilde{g}^{\prime}=f^{\prime}e^{-\tilde{g}}+fe^{-\tilde{g}}\frac{f^{\prime}}{f}=0.$

It follows that $f=Ce^{\tilde{g}}$ for some constant $C\neq 0$. Our desired
logarithm is then $g=\tilde{g}+\tilde{C}$, where $\tilde{C}$ is chosen so that
$e^{\tilde{C}}=C$.

###### Corollary 10

Let $f$ be an analytic function on a simply connected domain $W$, with
$f(z)\neq 0$ for all $z\in W$. Then there is an analytic $n$-th root of $f$ on
$W$; that is, there is an analytic function $g$ on $W$ such that $g^{n}=f$.

Proof: Choose $\tilde{g}$ so that $e^{\tilde{g}}=f$. The desired $n$-th root
is then $g=e^{\tilde{g}/n}$.

###### Corollary 11

Let $h$ be a (real valued) harmonic function on a simply connected domain $W$.
Then there is a harmonic conjugate of $h$ on $W$; that is, there is a harmonic
function $\tilde{h}$ on $W$ such that $h+i\tilde{h}$ is analytic.

Proof: Let us first note that if $h$ were the real part of an analytic
function $f$, then by the Cauchy-Riemann equations we could write
$f^{\prime}=h_{x}-ih_{y}$. We therefore begin by setting $g=h_{x}-ih_{y}$ and
letting $f$ be an antiderivative of $g$. If $f=u+iv$, then the Cauchy-Riemann
equations again show that $f^{\prime}=u_{x}-iu_{y}$, and it follows that
$h_{x}=u_{x},h_{y}=u_{y}$. From this we conclude that there is a constant $c$
such that $h+c=u$, and we see that we can take $\tilde{h}=v-ic$ to obtain the
conjugate to $h$.

###  4.2 Green’s function

In the field of analysis, Green’s function $G(x,y)$ on regions of
${\mathbb{R}}^{n}$ is formally defined to be the solution of
$LG(x,y)=\delta(y-x)$, where $L$ is a linear differential operator. In complex
analysis, where the Laplacian is the differential operator of most importance,
for a given domain $W\subseteq{\mathbb{C}}$ and $z\in W$ the Green’s function
of the Laplacian is generally defined by the following.

###### Definition 1

The Green’s function $G_{W}(z,w)$ on a domain $W$ is a function in $w$ on
$W\backslash\\{z\\}$ satisfying the following properties.

  * (i)

$G_{W}(z,w)$ is harmonic and positive on $W\backslash\\{z\\}$.

  * (ii)

$G_{W}(z,w)\longrightarrow 0$ as $w\longrightarrow bd(W)$, where $bd(W)$ is
the set-theoretic boundary of $W$ (Note that the boundary is to be taken in
the Riemann sphere $\hat{\mathbb{C}}:={\mathbb{C}}\cup\\{\infty\\}$, so that
if $W$ is unbounded then $\infty\in W$).

  * (iii)

$G_{W}(z,w)+\frac{1}{\pi}\ln|w-z|$ extends to be continuous (and therefore
harmonic) at $w=z$.

Note that the factor of $\frac{1}{\pi}$ in $(iii)$ is not standard but has
been chosen to align with the probabilistic considerations to follow. Not
every domain has a Green’s function as defined above, as for instance it can
be shown that no such function can exist on the punctured disk
${\mathbb{D}}^{\times}=\\{0<|z|<1\\}$, or more generally on a domain with
isolated singularities. The Green’s function is of tremendous importance in
analysis on ${\mathbb{R}}^{n}$, including complex analysis, and the question
of what domains possess a Green’s function has been keenly studied by analysts
over the years. As will be seen below, the existence of the Green’s function
on simply connected domains leads to a proof of the celebrated Riemann mapping
theorem.

We will now bring Brownian motion into the picture, which gives us an
advantage in that we may attach Green’s function to a stopping time rather
than to a domain, leading to a useful generalization. Let us begin by letting
$\rho_{t}^{T(W)}(z,w)$ be the probability density function at point $w$ and
time $t$ of a Brownian motion started at $z$ and stopped at the exit time
$T(W)$ of $W$. We then can calculate formally, for any suitable function $f$,

(4.10)
$\begin{split}E_{z}\int_{0}^{T(W)}f(B_{s})ds&=\int_{0}^{T(W)}E_{z}[f(B_{s})]ds\\\
&=\int_{0}^{\infty}\int_{W}f(w)\rho_{s}^{W}(z,w)dA(w)ds\\\
&=\int_{W}\Big{(}\int_{0}^{\infty}\rho_{s}^{W}(z,w)ds\Big{)}f(w)dA(w).\end{split}$

We see that the function

(4.11) $G_{T(W)}(z,w)=\int_{0}^{\infty}\rho_{s}^{W}(z,w)ds,$

is the density of the occupation measure
$\mu(A)=\int_{0}^{T(W)}1_{A}(B_{s})ds$. We will refer to (4.11) as the Green’s
function of $T(W)$, and show that it satisfies properties $(i)$ through
$(iii)$ above, but first let us note that the definition (4.11) makes sense if
$T(W)$ is replaced by an arbitrary stopping time $\tau$. Naturally, we will
denote the resulting function as $G_{\tau}(z,w)$. Our first order of business
is showing that $G_{\tau}(z,w)$ is conformally invariant, and subsequently
providing an extension of this statement to analytic functions which are not
necessarily injective.

###### Proposition 14

Green’s function is conformally invariant. That is, if $f$ is a conformal map
on $W$, then $G_{T(W)}(z,w)=G_{T(f(W))}(f(z),f(w))$.

Proof: As was described above, $G_{T(W)}(z,w)$ is the density of the
occupation measure $\mu_{z}(A)=\int_{0}^{T(W)}1_{A}(B_{s})ds$, where the
subscript of $z$ signifies that $B_{0}=z$ a.s. If $\lambda$ denotes Lebesgue
measure in the plane, then we have

(4.12) $G_{T(W)}(z,w)=\lim_{\delta\searrow
0}\frac{\mu_{z}(D(w,\delta))}{\lambda(D(w,\delta))};\qquad
G_{f(\Omega)}(f(z),f(w))=\lim_{\delta\searrow
0}\frac{\mu_{f(z)}(f(D(w,\delta)))}{\lambda(f(D(w,\delta)))}.$

Let $\varepsilon>0$ be given. Conformality implies that $f^{\prime}(w)\neq 0$,
and that for sufficiently small $\delta$ we have
$D(f(w),|f^{\prime}(w)|\delta(1-\varepsilon))\subseteq
f(D(w,\delta)))\subseteq D(f(w),|f^{\prime}(w)|\delta(1+\varepsilon))$, so
that
$\frac{\lambda(f(D(w,\delta)))}{\lambda(D(w,\delta))}\in(|f^{\prime}(w)|^{2}(1-\varepsilon)^{2},|f^{\prime}(w)|^{2}(1+\varepsilon)^{2})$.
On the other hand, in Lévy’s Theorem the scaling factor for time is
$|f^{\prime}|^{2}$ as well; that is, we can choose $\delta$ sufficiently small
so that
$\frac{|f^{\prime}(w^{\prime})|}{|f^{\prime}(w)|}\in(1-\varepsilon,1+\varepsilon)$
for all $w^{\prime}\in D(w,\delta)$, and then we will have
$\frac{\mu_{\Omega}(f(D(w,\delta)))}{\mu_{\Omega}(D(w,\delta))}\in(|f^{\prime}(w)|^{2}(1-\varepsilon),|f^{\prime}(w)|^{2}(1+\varepsilon))$.
Combining these two estimates and letting $\varepsilon\longrightarrow 0$ shows
that $\lim_{\delta\searrow
0}\frac{\mu_{\Omega}(D(w,\delta))}{\lambda(D(w,\delta))}=\lim_{\delta\searrow
0}\frac{\mu_{f(\Omega)}(f(D(w,\delta)))}{\lambda(f(D(w,\delta)))}$, and the
result follows.

It is important to note, however, that Lévy’s theorem in fact does not require
maps to be injective, permitting general nonconstant analytic functions as
well. This allows us to extend the previous proposition (with essentially the
same proof) as follows.

###### Theorem 22

Let $W$ be a domain, and suppose $f$ is a function analytic on $W$. Let
$B_{t}$ be a Brownian motion starting at $a$, and $\tau$ a stopping time such
that the set of Brownian paths $\\{B_{t}:0\leq t\leq\tau\\}$ lie within $W$
a.s. Let $\hat{\tau}=\sigma(\tau)$, where $\sigma$ is defined by L evy’s
Theorem. Then

(4.13) $G_{\hat{\tau}}(f(z),w)=\sum_{w^{\prime}\in
f^{-1}(\\{w\\})}n(f,w^{\prime})G_{\tau}(z,w^{\prime}),$

where $n(f,w^{\prime})$ is the order of the zero of $f-w$ at $w^{\prime}$.

The proof of the Proposition 14 applies to this more general statement at all
points which are not images under $f$ of a point $w^{\prime}$ at which
$f^{\prime}(w^{\prime})=0$ (the critical values of $f$), with the only
difference being that a point $w$ with multiple preimages will accumulate mass
in the projected Green’s function corresponding to mass accumulated at each of
the preimages by the initial Brownian motion. The $n(f,w^{\prime})$ term is
not necessary in order to calculate a density of the occupation measure, as in
fact $n(f,w^{\prime})=1$ except on the zero set of $f^{\prime}$, which is a
discrete set and therefore of Lebesque measure 0; the term is required merely
to make the right side of (4.13) continuous in $w$ when $G_{\tau}$ is as well
(such as when $\tau$ is an exit time of a domain), since if $w$ is a critical
value of $f$ then points near $w$ will have as many preimages as does $w$ only
when multiplicities of preimages are counted. For an example of this
phenomenon see Exercise ???.

Our eventual goal is to prove the Riemann mapping theorem in the next section,
and we must first calculate several of the simplest Green’s functions. We note
first the two-dimensional Gaussian density $\rho_{t}(z,w)=\frac{1}{2\pi
t}e^{-|z-w|^{2}/(2t)}$; it is evident that
$\int_{0}^{\infty}\rho_{t}(z,w)dt=\infty$ for any $z,w$, and thus when
$\tau=\infty$ we have $G_{\tau}(z,w)=\infty$.

As usual, let ${\mathbb{H}}=\\{Im(z)>0\\}$. The density
$\rho_{t}^{T({\mathbb{H}})}(z,w)$ must capture the probability of Brownian
paths near $w$ at time $t$, but only those which have not previously
intersected ${\mathbb{R}}$. By the reflection principle (ref ???), the
processes $B_{t}$ and

(4.14) $\hat{B}_{t}=\left\\{\begin{array}[]{ll}B_{t}&\qquad\mbox{if }t\leq
T({\mathbb{H}})\\\ \overline{B}_{t}&\qquad\mbox{if
}t>T({\mathbb{H}})\;,\end{array}\right.$

have the same law; but $B_{t}$ is near $w$ and $t>T({\mathbb{H}})$ precisely
when $\hat{B}_{t}$ is near $\bar{w}$ with $t>T({\mathbb{H}})$, and this occurs
precisely when $\hat{B}_{t}$ is near $\bar{w}$, since the Brownian motion
cannot travel from $z$ to $\bar{w}$ without first crossing ${\mathbb{R}}$. We
conclude that
$\rho_{t}^{T({\mathbb{H}})}(z,w)=\rho_{t}(z,w)-\rho_{t}(z,\bar{w})=\frac{1}{2\pi
t}(e^{-|z-w|^{2}/(2t)}-e^{-|z-\bar{w}|^{2}/(2t)})$. We therefore have

(4.15) $\begin{split}G_{T({\mathbb{H}})}(z,w)&=\int_{0}^{\infty}\frac{1}{2\pi
t}(e^{\frac{-|z-w|^{2}}{2t}}-e^{\frac{-|z-\bar{w}|^{2}}{2t}})dt\\\
&=\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{t}\int_{|z-w|}^{|z-\bar{w}|}\frac{a}{t}e^{\frac{-a^{2}}{2t}}dadt\\\
&=\frac{1}{2\pi}\int_{|z-w|}^{|z-\bar{w}|}\int_{0}^{\infty}\frac{a}{t^{2}}e^{\frac{-a^{2}}{2t}}dtda\\\
&=\frac{1}{\pi}\int_{|z-w|}^{|z-\bar{w}|}\frac{1}{a}\int_{0}^{\infty}e^{-u}duda\\\
&=\frac{1}{\pi}\ln\Big{(}\frac{|z-\bar{w}|}{|z-w|}\Big{)}.\end{split}$

Now, $\phi(z)=-i\Big{(}\frac{z-1}{z+1}\Big{)}$ maps ${\mathbb{D}}$ conformally
onto ${\mathbb{H}}$, sending $0$ to $i$. Thus,

(4.16)
$\begin{split}G_{T({\mathbb{D}})}(0,z)&=G_{T({\mathbb{H}})}(i,\phi(z))=\frac{1}{\pi}\ln\Big{(}\frac{|i-i(\frac{\bar{z}-1}{\bar{z}+1})|}{|i+i(\frac{z-1}{z+1})|}\Big{)}\\\
&=\frac{1}{\pi}\ln\Big{|}\frac{2z+2}{2|z|^{2}+2z}\Big{|}=\frac{1}{\pi}\ln\frac{1}{|z|}.\end{split}$

Let us now calculate the Green’s function of a stopping time defined by the
winding of Brownian motion. If our Brownian motion $B_{t}$ starts at 1 then it
will never hit 0 a.s., and we can then define $arg(B_{t})$ as a continuous
process (with $arg(B_{0})=0$). For any posi-

Figure 4.1: The stopping time $\tau_{1}$

tive integer $n$ we then define $\tau_{n}=\inf\\{t:arg(B_{t})=\pm 2\pi n\\}$.
In other words, $\tau_{n}$ is the first time at which the Brownian motion has
wound around the origin in either direction $n$ times. The image to the right
gives an intuitive illustration of a Brownian path up until time $\tau_{1}$.
If we let $U=\\{Re(z)>0\\}$ denote the right half-plane, we see that the
function $z\longrightarrow z^{4n}$ transforms a Brownian motion starting at
$1$ and stopped at $T(U)$ into one starting at 1 and stopped at $\tau_{n}$. We
may therefore calculate $G_{\tau_{n}}$ by using Proposition 22. For any point
$re^{i\theta}$, with $r>0$ and $\theta\in(-\pi,\pi]$, we have

$G_{\tau_{n}}(1,re^{i\theta})=\sum_{\begin{subarray}{c}k\in{\cal Z}\\\
(\frac{\theta}{4n}+\frac{2\pi
k}{4n})\in[-\frac{\pi}{2},\frac{\pi}{2}]\end{subarray}}G_{T(U)}(1,r^{\frac{1}{4n}}e^{i(\frac{\theta}{4n}+\frac{2\pi
k}{4n})}).$

The explicit formula is now easy to produce if desired, using the fact that
reflection across the $y$-axis is given by $z\longrightarrow-\bar{z}$, so that
(4.15) gives $G_{T(U)}(z,w)=\frac{1}{\pi}\ln\frac{|z+\bar{w}|}{|z-w|}$;
however, our purpose in examining this stopping time is the proof of the
Riemann mapping theorem in the next section, and for that only a few basic
properties of $G_{\tau_{n}}$ are required. To begin with, since $G_{\tau_{n}}$
is expressed as the sum of a finite number of finite terms, it is finite for
all $re^{i\theta}\neq 1$. Furthermore,
$G_{\tau_{n}}(1,re^{i\theta})+\frac{1}{\pi}\ln|1-re^{i\theta}|$ can be
extended continuously at $re^{i\theta}=1$, since the only term in the sum
which has a singularity corresponds to $k=0$ and has singular part
$-\frac{1}{\pi}\ln|1-r^{\frac{1}{4n}}e^{i\frac{\theta}{4n}}|$, and we have

(4.17) $\lim_{re^{i\theta}\longrightarrow
1}\frac{1}{\pi}\ln|1-re^{i\theta}|-\frac{1}{\pi}\ln|1-r^{\frac{1}{4n}}e^{i\frac{\theta}{4n}}|=\lim_{\zeta\longrightarrow
1}\ln\frac{1-\zeta^{4n}}{1-\zeta}=\ln 4n.$

Finally, if $r\longrightarrow 0$ then all of the preimages of $re^{i\theta}$
approach 0 as well, and we conclude that
$G_{\tau_{n}}(1,re^{i\theta})\longrightarrow 0$; an analogous argument shows
that $G_{\tau_{n}}(1,re^{i\theta})\longrightarrow 0$ whenever
$r\longrightarrow\infty$.

###  4.3 Conformal mapping and the Reimann mapping theorem

The Riemann mapping theorem is as follows.

Riemann mapping theorem If $W\subset{\mathbb{C}}$ is a simply connected domain
with $W\neq{\mathbb{C}}$, and $a\in\Omega$, then there is a conformal map $f$
from $W$ onto ${\mathbb{D}}$, with $f(a)=0$.

Note that if such a map $f$ existed, then the function
$\frac{-1}{\pi}\ln|f(z)|$ would be satisfy the conditions $(i)-(iii)$ given
for the Green’s function in the previous section, and thus would be equal to
$G_{W}(a,z)$. We have already seen that the Green’s function exists for $W$,
and therefore the trick is to use $G_{U}$ to construct $f$.

###  4.4 Schwarz-Christoffel transformations

Another important class of conformal maps are the Schwarz-Christoffel
transformations. In the simplest form, this maps the upper half-plane
${\mathbb{H}}=\\{Im(z)>0\\}$ conformally onto a polygon. It is easiest to
understand this type of map by looking first at the map $f(z)=z^{m}$, for
$0<m\leq 2$, which takes ${\mathbb{H}}$ conformally to the domain
$\\{0<Arg(z)<m\pi\\}$. We see that this function turns an angle of $\pi$ (a
straight angle) into an angle of $m\pi$. Note also that the derivative, up to
a constant, is $z^{m-1}$.

Suppose now that we want to find a conformal map $f$ from ${\mathbb{H}}$ onto
a polygon with vertices $w_{1},\ldots,w_{n}$ with corresponding angles (to the
left) of $\alpha_{1}\pi,\ldots,\alpha_{n}\pi$. Such a conformal map will
extend to a bijection between the boundaries of the domains, and there will
therefore be points $z_{1},\ldots,z_{n-1}$ on ${\mathbb{R}}$ which are mapped
to $w_{1},\ldots,w_{n-1}$, and we will arrange it so that $\infty$ is mapped
to $w_{n}$. Between any two of these points, $z_{j}$ and $z_{j+1}$ say, the
argument of $f^{\prime}(z)$ must remain constant, but then it must have a jump
of $\alpha_{j}\pi$ at $z_{j}$. This suggests that

$f^{\prime}(z)=C(z-z_{1})^{\alpha_{1}-1}(z-z_{2})^{\alpha_{2}-1}\ldots(z-z_{n-1})^{\alpha_{2}-1}.$

This turns out to be correct, and we have the following theorem.

###### Theorem 23

The conformal map $f$ from ${\mathbb{H}}$ onto a polygon with vertices
$w_{1},\ldots,w_{n}$ with corresponding angles (to the left) of
$\alpha_{1}\pi,\ldots,\alpha_{n}\pi$ is given by

$f(z)=A+C\int^{z}(\zeta-z_{1})^{\alpha_{1}-1}(\zeta-
z_{2})^{\alpha_{2}-1}\ldots(\zeta-z_{n-1})^{\alpha_{2}-1}d\zeta,$

for some complex constants $A,C$, where $w_{j}=f(z_{j})$ for $j=1,\ldots,n-1$.

Most explicit conformal maps that are known can be expressed as compositions
of Möbius transformations and Schwarz-Christoffel transformations. On the
other hand, the class of conformal maps is in fact very large (as is shown by
the Riemann Mapping Theorem below), we just don’t have formulas for most of
them.

###  4.5 Exercises

1 a) Find all entire functions $f(z)$ such that $f(1/n)=1/n^{2}$ for all
$n\in{\mathbb{N}}$.

b) Find all entire functions $f(z)$ such that $f(1/n)=1/\sqrt{n}$ for all
$n\in{\mathbb{N}}$.

2 Suppose $f(z)$ is an analytic function on a domain $W$ with the property
that $f(z)^{n}\in{\mathbb{R}}$ for all $z\in W$. Show that $f$ is a constant
function.

3 Suppose that $\gamma$ is a Jordan curve: that is, $\gamma$ can be realized
as the image of a continuous and one-to-one function $\gamma(t)$ from $[0,1]$
into ${\mathbb{C}}$. $\gamma$ then determines a bounded domain $W$ in its
interior. Suppose that $f(z)$ is an analytic function on $W$, and suppose that
there are points $w_{1},\ldots,w_{n}\in\gamma$ such that
$\limsup_{\\{z\longrightarrow\delta
W\backslash\\{w_{1},\ldots,w_{n}\\}\\}}|f(z)|\leq m<\infty$. Suppose further
that there is some finite $M>m$ such that $|f(z)|\leq M$ for all $z\in W$.
Show that in fact $|f(z)|\leq m$ for all $z\in W$.

4 Let $W$ be the square domain $\\{-1<x<1,-1<y<1\\}$. Suppose $f(z)$ is a
bounded analytic function on $W$ such that $f(z)\longrightarrow 0$ as $z$
approaches the bottom edge of the square; that is,
$\lim_{\\{z\longrightarrow\\{y=-1\\}\\}}|f(z)|=0$. Prove that $f(z)=0$ for all
$z\in W$.

5 Suppose $p(z)=\beta_{k}z^{k}+\ldots+\beta_{0}$ is a degree $k$ polynomial,
with $\beta_{0},\ldots,\beta_{k}\in{\mathbb{C}}$ and $\beta_{k}\neq 0$. Show
that $p$ has $k$ roots, counting multiplicities. That is, show that there are
values $a_{1},\ldots,a_{k},c\in{\mathbb{C}}$ such that
$p(z)=c(z-a_{1})(z-a_{2})\ldots(z-a_{k})$. (Hint: If $p(a)=0$, consider
$p(z)-p(a)$).

6 i) (Independence of parameterizations) Suppose
$\gamma_{1}:[a_{1},b_{1}]\longrightarrow{\mathbb{C}}$ and
$\gamma_{2}:[a_{2},b_{2}]\longrightarrow{\mathbb{C}}$ are rectifiable curves,
and suppose that there is a smooth increasing function
$\phi:[a_{1},b_{1}]\longrightarrow[a_{2},b_{2}]$ with
$\phi(a_{1})=a_{2},\phi(b_{1})=b_{2}$ such that
$\gamma_{1}(t)=\gamma_{2}(\phi(t))$ for all $t\in[a_{1},b_{1}]$. Show that

(4.18) $\int_{\gamma_{1}}f(z)dz=\int_{\gamma_{2}}f(z)dz$

for any analytic function $f$ defined in a neighborhood of
$\gamma_{1}=\gamma_{2}$.

ii) (Path reversal) Let $\gamma:[a_{1},b_{1}]\longrightarrow{\mathbb{C}}$ be a
rectifiable curve. Define
$\tilde{\gamma}:[-b_{1},-a_{1}]\longrightarrow{\mathbb{C}}$ by
$\tilde{\gamma}(t)=\gamma(-t)$. Then $\tilde{\gamma}$ is simply the curve
$\gamma$ traversed in the opposite direction. Show that

(4.19) $\int_{\tilde{\gamma}}f(z)dz=-\int_{\gamma}f(z)dz$

for any analytic function $f$ defined in a neighborhood of $\gamma$.

iii) (Concatenating paths) Suppose
$\gamma_{1}:[a,b]\longrightarrow{\mathbb{C}}$ and
$\gamma_{2}:[c,d]\longrightarrow{\mathbb{C}}$ are rectifiable curves such that
$\gamma_{1}(b)=\gamma_{2}(c)$. Define a new curve
$\tilde{\gamma}:[a,b+d-c]\longrightarrow{\mathbb{C}}$ by

(4.20) $\tilde{\gamma}(t):=\begin{cases}\gamma_{t}(t)&t\in[a,b],\\\
\gamma_{2}(t-b+c)&t\in[b,b+d-c].\end{cases}$

$\tilde{\gamma}$ is the concatenation of $\gamma_{1}$ and $\gamma_{2}$. Show
that

(4.21)
$\int_{\tilde{\gamma}}f(z)dz=\int_{\gamma_{1}}f(z)dz+\int_{\gamma_{2}}f(z)dz$

for any analytic function $f$ defined in a neighborhood of $\tilde{\gamma}$.

7 Let $d(z,\delta W)$ be as in Section 4.1. Show that the function
$z\longrightarrow d(z,\delta W)$ is continuous and positive on $W$. Conclude
that if $K$ is a compact set in $W$ then there is some $\varepsilon>0$ such
that $d(z,\delta W)>0$ for all $z\in K$.

8 Show that the radius of convergence is the same for the three series
$\sum_{n=0}^{\infty}\beta_{n}(z-z_{0})^{n},\sum_{n=0}^{\infty}\frac{\beta_{n}}{n+1}(z-z_{0})^{n+1},$
and $\sum_{n=1}^{\infty}n\beta_{n}(z-z_{0})^{n-1}$.

9 Polygon admits triangulation.

1. Let $\gamma=\\{e^{it}:0\leq t\leq 2\pi\\}$. Evaluate the following two integrals.

(4.22) $\int_{\gamma}\frac{1}{z}dz,\qquad\int_{\gamma}\frac{1}{z^{2}}dz.$

2. Let $\gamma(t):[0,5\pi]$ be defined as follows.

(4.23) $\gamma(t)=\begin{cases}(t+\frac{3\pi}{2})e^{it}&0\leq t\leq 2\pi,\\\
\frac{3\pi}{2}+2\pi e^{-it}&2\pi\leq t\leq 3\pi,\\\
(t-\frac{7\pi}{2})+i\cos(t-\frac{7\pi}{2})&3\pi\leq t\leq 5\pi.\end{cases}$

Sketch $\gamma$, and evaluate

(4.24) $\int_{\gamma}e^{1/z}dz.$

3. Suppose $p(z)=\beta_{k}z^{k}+\ldots+\beta_{0}$ is a degree $k$ polynomial, with $\beta_{0},\ldots,\beta_{k}\in{\mathbb{C}}$ and $\beta_{k}\neq 0$. Show that $p$ has $k$ roots, counting multiplicities. That is, show that there are values $a_{1},\ldots,a_{k},c\in{\mathbb{C}}$ such that $p(z)=c(z-a_{1})(z-a_{2})\ldots(z-a_{k})$. (Hint: If $p(a)=0$, consider $p(z)-p(a)$).

10 Suppose $W$ is a domain and $C$ is a line intersecting $W$. Suppose further
that $f$ is a continuous function on $W$ which is analytic on $W\backslash C$.
Show that in fact $f$ is analytic on all of $W$. (Hint: Morera’s Theorem)

##  Chapter 5 The reflection principle and consequences

In this chapter we study a number of important consequences of the reflection
principle for Brownian motion. Let $T_{\mathbb{R}}$ be the hitting time of
${\mathbb{R}}$, and set

(5.1) $\hat{B}_{t}=\begin{cases}B_{t}&t\leq T_{\mathbb{R}},\\\
\stackrel{{\scriptstyle\line(1,0){10.0}}}{{B_{t}}}&t>T_{\mathbb{R}}.\end{cases}$

The reflection principle for Brownian motion says that $\hat{B}_{t}$ is itself
a Brownian motion.

###  5.1 The reflection principle for harmonic functions

The reflection principle for Brownian motion leads immediately to the
reflection principles for harmonic functions.

###### Theorem 24

Suppose that $h^{+}(z)$ is harmonic on a domain $W^{+}\subseteq{\mathbb{H}}$.
Suppose further that $h^{+}(z)\longrightarrow 0$ as $Im(z)\longrightarrow 0$
in $W^{+}$. Let $W^{-}=\\{\bar{z}:z\in W^{+}\\}$ be the reflection of $W^{+}$
over ${\mathbb{R}}$, and let $W=W^{+}\cup W^{-}\cup(\delta
W^{+}\cap{\mathbb{R}})^{o}$, where $(\delta W^{+}\cap{\mathbb{R}})^{o}$ is the
interior of the set $(\delta W^{+}\cap{\mathbb{R}})$, taken as a subset of
${\mathbb{R}}$. For $z\in W$, define

(5.2) $h(z)=\begin{cases}h^{+}(z)&z\in W^{+},\\\ -h^{+}(\bar{z})&z\in
W^{-},\\\ 0&z\in(\delta W^{+}\cap{\mathbb{R}})^{o}.\end{cases}$

Then $h(z)$ is harmonic on $W$.

Remark: Note that it is not being claimed that $W$ is a domain, since it will
not be connected if $(\delta W^{+}\cap{\mathbb{R}})^{o}=\emptyset$. However,
$W$ is a domain if $(\delta W^{+}\cap{\mathbb{R}})^{o}\neq\emptyset$, and an
open set in either case.

Proof: Harmonicity is a local property, so it suffices to consider $z$ in the
various regions of $W$ separately. Clearly, $h$ is harmonic on $W^{+}$. If
$z_{0}\in W^{-}$, we know from Theorem 6 that $h^{+}$ can be expanded into a
power series in a neighborhood of
$\stackrel{{\scriptstyle\line(1,0){8.0}}}{{z_{0}}}$, so that we may write

(5.3)
$\begin{split}h(z)=-h^{+}(\bar{z})&=-\Big{(}\beta_{0}+\sum_{n=1}^{\infty}\beta_{-n}\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(\bar{z}-\stackrel{{\scriptstyle\line(1,0){8.0}}}{{z_{0}}})}}^{n}+\sum_{n=1}^{\infty}\beta_{n}(\bar{z}-\stackrel{{\scriptstyle\line(1,0){8.0}}}{{z_{0}}})^{n}\Big{)}\\\
&=-\beta_{0}+\sum_{n=1}^{\infty}(-\beta_{-n})(z-z_{0})^{n}+\sum_{n=1}^{\infty}(-\beta_{n})\stackrel{{\scriptstyle\line(1,0){30.0}}}{{(z-z_{0})}}^{n}.\end{split}$

This shows that $h$ is harmonic in a neighborhood of $z_{0}$. Finally, if
$z_{0}\in(\delta W^{+}\cap{\mathbb{R}})^{o}$, then we can choose $r>0$ such
that $\bar{D}(z_{0},r)\subseteq W$. Let $D=D(z_{0},r)$. If we can show that
$E_{z}[h(B_{T_{D}})]=h(z)$ for any $z\in D$, then by Proposition 11 it will
follow that $h$ is harmonic on $D$, and in particular at $z_{0}$. If
$z\in(D\cap{\mathbb{R}})$, then by the symmetry of $h$ it is clear that
$E_{z}[h(B_{T_{D}})]=0=h(z)$. Suppose now that $z\in(W^{+}\cap D)$. Let
$D^{+}=(W^{+}\cap D)$, and write

(5.4)
$E_{z}[h(B_{T_{D}})]=E_{z}[h(B_{T_{D^{+}}})1_{\\{T_{D^{+}}<T_{D}\\}}]+E_{z}[h(B_{T_{D^{+}}})1_{\\{T_{D^{+}}\geq
T_{D}\\}}].$

Note that, since $\hat{B}_{t}$ is itself a Brownian motion, we must have