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ABSTRACT

Zero Distribution of Generated Taylor Polynomials

Our goal in this paper is to study the zero distribution of a sequence of
polynomials whose coefficients satisfy a three-term recurrence. Equivalently,
these polynomials are Taylor polynomials of a rational function with a
polynomial denominator of degree 2. We will use complex analysis to find the
bi-variate generating function for that sequence of Taylor polynomials. With
this generating function, we prove that the zeros of these Taylor polynomials
lie on one side of the closed disk centered at the origin. The radius of the
disk is exactly the reciprocal of the modulus of the smaller zero of the
degree two denominator. Finally, we show that the zeros of these Taylor
polynomials approach the boundary of the disk above.

Juhoon Chung

May 2022

Zero Distribution of Generated Taylor Polynomials

by

Juhoon Chung

A project

submitted in partial

fulfillment of the requirements for the degree of

Master of Science in Mathematics

in the College of Science and Mathematics

California State University, Fresno

May 2022

APPROVED

For the Department of Mathematics:

> We, the undersigned, certify that the project of the following student meets
> the required standards of scholarship, format, and style of the university
> and the student’s graduate degree program for the awarding of the master’s
> degree.

Juhoon Chung  Project Author

Dr. Khang Tran Mathematics

Dr. Stefaan Delcroix Mathematics

Dr. Oscar Vega Mathematics

ACKNOWLEDGMENTS

I do not know how many times my parents were disappointed by me. What I can
tell is that they always try to encourage me and cheer me. Throughout working
with Dr. Tran, I did not catch up his expectation. He still keeps his respect
upon me, and helps me. I appreciate his patience. I would like to thank
Dr.Tran for his intuitive conjecture and skills for this project.

I have felt the passion for their work from every professors I have been
taught by ever since I entered Fresno State. I felt their diligent behavior to
their works make me a little humble and push myself to make effort.

###### TABLE OF CONTENTS

  1. INTRODUCTION
  2. COMPLEX ANALYSIS
    1. CONSEQUENCES OF THEOREMS ABOUT ZERO DISTRIBUTIONS
  3. POLYNOMIALS GENERATED BY $\displaystyle{\frac{1}{(at^{2}+bt+c)(1-tz)}}$
    1. ZERO DISTRIBUTION OF $\\{P_{m}(z)\\}_{m\geq 0}$ ABOUT A CIRCLE
  4. APPLICATION OF LIMIT DISTRIBUTION OF ZEROS
    1. CONCLUSIONS
  5. REFERENCES

###### LIST OF FIGURES

  1. 0.1  Zeros of Taylor polynomials of $e^{z}$ for $N=15,30,50$
  2. 0.2  Zeros inside of the corresponding circles
  3. 0.3  Zeros of $P_{m}(z)$ for $m=15,30,100$
  4. 0.4  Zeros of $f_{m}(z)$ for $m=10,15,50$

## INTRODUCTION

Many famous sequences of numbers in mathematics, such as the Fibonacci
numbers, satisfy recurrent relations. An equivalent way to represent those
relations is to use ordinary generating functions. For example, it is known
that the sequence of Fibonacci numbers has the generating function (see [10])

$\frac{t}{1-t-t^{2}}.$

For a rigorous definition of generating function, we refer the reader to the
section ‘Polynomials Generated by $\frac{1}{(at^{2}+bt+c)(1-tz)}$’. Similarly,
we can define a recursive sequence of functions from a bi-variate generating
function. For example, the famous sequence of Chebyshev polynomials,
$U_{m}(z)$, satisfying the recurrence

$\displaystyle U_{0}(z)=$ $\displaystyle 1$ $\displaystyle U_{1}(z)=$
$\displaystyle 2z$ $\displaystyle U_{n+1}(z)=$ $\displaystyle
2zU_{n}(z)-U_{n-1}(z),n\geq 1,$

has the generating function (see [11])

$\sum_{m=0}^{\infty}U_{m}(z)t^{m}=\frac{1}{1-2zt+t^{2}}.$

A question may occur: “Is there a pattern in the zeros of such a sequence of
functions?” Many studies have taken place to answer this question.

In 2017, Forgács and Tran in [8] found sufficient conditions of $P(t)$ and $r$
for $\\{H_{m}(z)\\}_{m\geq 0}$, for large $m$, generated by

$\frac{1}{P(t)+zt^{r}},$

to have zeros inside an interval $(a,b)$. They found explicit $a$ and $b$ such
that, if we denote $\mathcal{Z}(H_{m})$ the set of all the zeros of
$H_{m}(z)$, $\bigcup_{m\in\mathbb{M}}\mathcal{Z}(H_{m})$ is dense in $(a,b)$.

In 2020, Forgács and Tran in [9], for $P(t)$ and $Q(t)$ polynomials with
positive real zeros, found sufficient condition of $r$ and polynomials
$P(t),Q(t)$ for $\\{H_{m}(z)\\}_{m\geq 0}$, generated by

$\frac{1}{P(t)+zt^{r}Q(t)},$

to have real zeros of the same sign, for all $m\in\mathbb{N}$. They showed
that $\bigcup_{m\geq 0}\mathcal{Z}(H_{m})$ is dense in an explicit real ray.

In 2018, Tran and Zumba in [7] found that the zeros of polynomials
$\\{H_{m}(z)\\}_{m\geq 0}$ generated by

$\frac{1}{1+ct+bt^{2}+zt^{3}}$

are real, for all $m\in\mathbb{N}$, if and only if one of the following
conditions holds:

  1. 1.

$c=0$ and $b\geq 0$

  2. 2.

$c\neq 0$ and $-1\leq\frac{b}{c^{2}}\leq\frac{1}{3}$.

If the first condition 1 holds, the union of all the zeros $\bigcup_{m\geq
0}\mathcal{Z}(H_{m})$ is dense on $(-\infty,\infty)$. If the second condition
2 holds, $\bigcup_{m\geq 0}\mathcal{Z}(H_{m})$ is dense on an explicit
interval depending on $b$ and $c$.

In 2020, Tran and Zumba found necessary and sufficient conditions for the
reality of the zeros of the sequence of polynomials generated by

$\frac{1}{1+at+(b+cz)t^{2}+(d+ez)t^{3}},$

where $a,b,c,d,e\geq 0$ and $ae\neq 0$ (see [6]). In the same paper, the
authors also found an optimal interval containing the zeros of the generated
sequence. Here, the optimality means that the union of all the zeros of the
generated sequence forms a dense subset of the given interval.

In this paper, we will study some results similar to those of the above
papers. Then we will make some new contribution to study the zero distribution
of polynomials. Our contribution would be related to Taylor polynomials, which
are the partial sums of the Taylor series. For example, from the famous Taylor
series for $e^{x}$

$e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!},$

we can define the sequence of Taylor polynomials $\\{P_{n}(x)\\}_{n\geq 0}$ as

$P_{n}(x)=\sum_{k=0}^{n}\frac{x^{k}}{k!}.$

Figure 0.1:  Zeros of Taylor polynomials of $e^{z}$ for $N=15,30,50$

In 1924, Szegö studied the zero distribution of the Taylor polynomials of
$e^{z}$ (see [12]). Figure 0.1 shows those zeros are about to form a curve.

In this paper, we use generating functions to study Taylor polynomials. For
example,

$\frac{e^{t}}{1-zt}$

is the generating function of the sequence of the reciprocal of the Taylor
Polynomials of $e^{z}$. The reader may found reasoning for this generating
function in Section ‘Complex Analysis’.

Our main goal is to study the zero distribution of a Taylor polynomial
generated by

$\frac{1}{at^{2}+bt+c}.$

In particular, we prove the following results.

######  Theorem 1.

Let $a,b,c\in\mathbb{R}$, where $ac<0$ and $b\neq 0$. Suppose
$\\{P_{m}(z)\\}_{m\geq 0}$ is a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then, for all $m\geq 0$, all the zeros of $P_{m}(z)$ lie on the closed disk
centered at the origin with radius $1/|\alpha|$ where $\alpha$ is the smallest
(in modulus) zero of $at^{2}+bt+c$.

######  Theorem 2.

Let $a,b,c\in\mathbb{R}$ where $ac>0$ and $b^{2}-4ac>0$. Suppose
$\\{P_{m}(z)\\}_{m\geq 0}$ is a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then, for all $m\geq 0$, no zero of $P_{m}(z)$ lie inside the closed disk
centered at the origin with radius $1/|\alpha|$ where $\alpha$ is the smallest
(in modulus) zero of $at^{2}+bt+c$.

Finally, we prove the following theorem about limiting behavior of the zero
distribution.

######  Theorem 3.

Let $a,b,c\in\mathbb{R}\backslash\\{0\\}$ where $b^{2}-4ac>0$. Suppose
$\\{P_{m}(z)\\}_{m\geq 0}$ is a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then

$\lim\mathcal{Z}(P_{m})=\left\\{z\in\mathbb{C}:|z|=\frac{1}{|\alpha|}\right\\}$

where $\alpha$ is the smallest (in modulus) zero of $at^{2}+bt+c$.

For the formal definition of $\lim\mathcal{Z}(P_{m})$, we refer the reader to
Section ‘Application of Limit Distribution of Zeros’.

The paper is organized as follows.

  1. 1.

Review concepts in Complex Analysis as a basis for the following sections;

  2. 2.

Study previous theorems relative to the zero distribution of the generated
sequence of functions;

  3. 3.

Applying skills learned above and with our methods to prove two main results.

The two main results are the followings

  * •

the zero distribution of Taylor Polynomials of rational polynomials on one
side of the circle (c.f. Theorem 1, Theorem 2),

  * •

the limiting behavior of the zeros of Taylor polynomials to the circle (c.f.
Theorem 3).

We will prove those two main results in the last two sections, respectively.

## COMPLEX ANALYSIS

In most of theorems from this paper, we will use generating functions to study
the zeros of a generated sequence of functions. We will develop rigorous
definitions of a bi-variate generating function. In order to achieve this
goal, we will review basic complex analysis. We start with the definition of
the radius of convergence of a power series.

######  Theorem 4 (Proposition 1.3, [2]).

For a given power series $\sum_{n=0}^{\infty}a_{n}(z-a)^{n}$, define the
extended real number $R$, $0\leq R\leq\infty$, by

$\frac{1}{R}=\lim\sup|a_{n}|^{1/n}.$

Then

  1. 1.

if $|z-a|<R$, the series converges absolutely;

  2. 2.

if $|z-a|>R$, the terms of the series become unbounded and so the series
diverges;

  3. 3.

if $0<r<R$ then the series converges uniformly on $\\{z;|z|\leq r\\}$.

Moreover, $R$ is unique having properties 1 and 1.

The extended real number $R$ is called the radius of convergence of the power
series.

Many of the following theorems are relative to ‘analytic functions’, defined
as follows.

######  Definition 5 (Definition 2.3, [2]).

Let $G$ be an open connected set. A function $f:G\rightarrow\mathbb{C}$ is
analytic if $f$ is continuously differentiable.

This paper studies theorems related to sequence of functions. In several
theorems, an explicit form of sequence of functions is unknown. So we must
derive it from the given generating function. We will find the region where
the given generating function is analytic. Then with the following theorem we
can transform the generating function to an infinite series.

######  Theorem 6 (Theorem 2.8, [2]).

Let $f$ be analytic on $B(R;a)$; then
$f(z)={\displaystyle\sum_{n=0}^{\infty}a_{n}(z-a)^{n}}$ for $|z-a|<R$ where
$a_{n}=\frac{1}{n!}f^{(n)}(a)$ and this series has radius of convergence
greater than or equal to $R$.

If we consider bi-variate function and build infinite series about one
variable, the coefficient would depend on the other variable. So we may
consider it as a function about the other variable. So we shall define the
generated sequence of functions as follows.

######  Definition 7.

For $z\in\mathbb{C}$, assume that $f(t,z)$ is analytic in $t$ in a
neighborhood around the origin. If $P_{m}(z)$ is the $t^{m}$-coefficient of
$f(t,z)$ expanded as a power series in $t$, then we say the sequence
$\\{P_{m}(z)\\}$ is generated by $f(t,z)$.

Although Theorem 6 provide us a way to obtain the $t^{m}$-coefficient as
follows

$P_{m}(z)=\frac{1}{m!}\frac{\partial f^{m}}{\partial^{m}t}(0,z),$

we will not use this method to obtain $\\{P_{m}(z)\\}_{m\geq 0}$. In all the
three main theorems, we will work with the rational polynomial

$f(z,t)=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

The partial fractions will transform the generating function $f(z,t)$ into the
form

$\frac{1}{1-pt}+\frac{1}{1-qt}+\frac{1}{1-zt}.$

for some $p,q\in\mathbb{R}$. Then we expand each fraction as follows

$\sum_{k=0}^{\infty}(at)^{k}+\sum_{k=0}^{\infty}(bt)^{k}+\sum_{k=0}^{\infty}(zt)^{k}.$

Then we will combine those three series into one series. Before combine
series, we need to verify if they are absolutely convergent by checking their
radius of convergence. The following theorem allow us to combine these series.

######  Proposition 8 (Proposition 1.6, [2]).

Let $\sum_{n=0}^{\infty}a_{n}(z-a)^{n}$ and
$\sum_{n=0}^{\infty}b_{n}(z-a)^{n}$ be power series with radius of convergence
greater than or equal to $r>0$. Put

$c_{n}=\sum_{k=0}^{n}a_{k}b_{n-k};$

Then both power series $\sum_{n=0}^{\infty}(a_{n}+b_{n})(z-a)^{n}$ and
$\sum_{n=0}^{\infty}c_{n}(z-a)^{n}$ have radius of convergence greater than or
equal to $r$, and

(0.1) $\displaystyle\sum_{n=0}^{\infty}(a_{n}+b_{n})(z-a)^{n}$
$\displaystyle=\left[\sum_{n=0}^{\infty}a_{n}(z-a)^{n}+\sum_{n=0}^{\infty}b_{n}(z-a)^{n}\right]$
$\displaystyle\sum_{n=0}^{\infty}c_{n}(z-a)^{n}$
$\displaystyle=\left[\sum_{n=0}^{\infty}a_{n}(z-a)^{n}\right]\left[\sum_{n=0}^{\infty}b_{n}(z-a)^{n}\right]$

for $|z-a|<r$.

So we can combine series into one series by grouping $t^{m}$-coefficients as
Equation (0.1) shows.

######  Proposition 9 (Proposition 2.5, [2]).

Let $f(z)={\displaystyle\sum_{n=0}^{\infty}a_{n}(z-a)^{n}}$ have radius of
convergence $R>0$. Then

  1. 1.

For each $k\geq 1$ the series

(0.2) $\sum_{n=0}^{\infty}n(n-1)\cdots(n-k+1)a_{n}(z-a)^{n-k}$

has radius of convergence $R$;

  2. 2.

The function $f$ is infinitely differentiable on $B(R;a)$ and, furthermore,
$f^{(k)}(z)$ is given by the series (0.2), for all $k\geq 1$ and $|z-a|<R$;

  3. 3.

For $n\geq 0$,

$a_{n}=\frac{1}{n!}f^{(n)}(a).$

In this paper, we study relations between generated sequence of functions and
recurrence relation, and the consequence 3 of Proposition 9 confirm the
uniqueness of coefficient. For example, if we are given a relation

$1=\sum_{k=0}^{\infty}[f_{k}(z))+zf_{k+1}(z)]t^{k}$

we consider both sides as a series. For the left hand side we consider it as a
series with 0 coefficient for all $t^{k}$, $k=1,2,\dots$, except for $k=0$. By
uniqueness of coefficient, if we compare both sides’ coefficient, we obtain as
follows

$\displaystyle f_{0}(z)+zf_{1}(z)=$ $\displaystyle 1$ $\displaystyle
f_{1}(z)+zf_{2}(z)=$ $\displaystyle 0$ $\displaystyle f_{2}(z)+zf_{3}(z)=$
$\displaystyle 0$ $\displaystyle\vdots$

which yields recurrence relation

$f_{m}(z)+zf_{m+1}(z)=0$

Our Proposition 9 allows us to prove the following proposition. The following
statement of Proposition 10 is a little different from that of Corollary 2.9
in [2]. We can deduce that the following Proposition 10 and Corollary 2.9 in
[2] are equivalent by Theorem 4.

######  Proposition 10.

(Corollary 2.9 [2]) Suppose that $\sum_{n=0}^{\infty}a_{n}z^{n}$ is absolutely
convergent on $B(R,0)$. Then $\sum_{n=0}^{\infty}a_{n}z^{n}$ is analytic on
$B(R,0)$.

###### Proof.

Let $r$ be the radius of convergence of $\sum_{n=0}^{\infty}a_{n}z^{n}$. First
we will prove that $r\geq R$. Suppose, by contrary, that $r<R$. Notice that
$B(R,0)\backslash B(r,0)$ is not empty. If we choose $z_{0}\in
B(R,0)\backslash B(r,0)$ arbitrarily, we are given in the hypothesis that
$\sum_{n=0}^{\infty}a_{n}z_{0}^{n}$ converges absolutely. Meanwhile, by
Theorem 4, $\sum_{n=0}^{\infty}a_{n}z_{0}^{n}$ diverges, a contradiction. Thus
$r\geq R$.

Moreover, Proposition 9 induces that $\sum_{n=0}^{\infty}a_{n}z^{n}$ is
continuously differentiable on $B(r,0)$. That is, analytic on $B(r,0)$.
Consequently, $\sum_{n=0}^{\infty}a_{n}z^{n}$ is analytic on $B(R,0)$. ∎

If we combine a couple of series into one seires, by Proposition 8, the
combined series is still absolutely convergent. Then above Proposition 10
comfirms that the combined series is analytic.

All the main theorems in this paper are about the reciprocal of a Taylor
polynomial. We shall define what a Taylor polynomial is, and what is its
reciprocal polynomial.

######  Definition 11 (p118, [3]).

We shall call

$P_{n}(z)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(z-a)^{k}$

the Taylor Polynomial of order $n$ generated by $f$ centered at $a$.

######  Definition 12.

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a polynomial

$f(t)=a_{0}+a_{1}t+\dots+a_{n}t^{n}$

with $a_{n}\neq 0$. Then

$g(t)=a_{n}+a_{n-1}t+\dots+a_{1}t^{n-1}+a_{0}t^{n}=t^{n}f\left(\frac{1}{t}\right)$

is the reciprocal polynomial of $f$.

In the first page of this paper, we have seen our generating function

$\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Next, we will obtain the relation between a generating function and the
corresponding reciprocal Taylor polynomial. To study the relation, we need to
reorder the addends of series so we would review the following theorem.

######  Theorem 13 (Theorem 6.27, [3]).

If $\sum_{k=1}^{\infty}a_{k}$ converges absolutely, and
$\sum_{k=1}^{\infty}b_{k}$ is any rearrangement of $\sum_{k=1}^{\infty}a_{k}$,
then $\sum_{k=1}^{\infty}b_{k}$ converges and

$\sum_{k=1}^{\infty}a_{k}=\sum_{k=1}^{\infty}b_{k}.$

We shall read the theorem of relation between the generating function and the
reciprocal Taylor polynomial.

######  Theorem 14.

The function $\frac{f(t)}{1-tz}$ is the generating function of the reciprocal
polynomial of Taylor Polynomials generated by $f(t)$ centered at $0$.

###### Proof.

Let

$f(z)=\sum_{m=0}^{\infty}a_{m}z^{m}$

be the Taylor Series of $f$. Define

$T_{n}(z):=\sum_{m=0}^{n}a_{m}z^{m}$

be the Taylor Polynomial of $f$ centered at $0$. Let $z\neq 0$ be arbitrary.
Suppose that $r$ is the radius of convergence of
$\sum_{m=0}^{\infty}T_{m}(z)t^{m}$. Observe that, for arbitrary $t$ such that
$|t|<\max(r,1)$, the series is absolutely convergent and thus, by Theorem 13,
we may rearrange the order of the series as follows:

$\displaystyle\sum_{m=0}^{\infty}T_{m}(z)t^{m}$ $\displaystyle=$
$\displaystyle a_{0}+(a_{0}+a_{1}z)t+(a_{0}+a_{1}z+a_{2}z^{2})t^{2}\cdots$
$\displaystyle=$ $\displaystyle
a_{0}(1+t+t^{2}+\cdots)+a_{1}z(t+t^{2}+t^{3}+\cdots)+a_{2}z^{2}(t^{2}+t^{3}+\dots)+\dots$

Since $|t|<\max(r,1)\leq 1$, in view of geometric series, it holds that

$\displaystyle 1+t+t^{2}+\cdots=$ $\displaystyle\frac{1}{1-t}.$

This implies that

$\displaystyle
a_{0}(1+t+t^{2}+\cdots)+a_{1}z(t+t^{2}+t^{3}+\cdots)+a_{2}z(t^{2}+t^{3}+\cdots)+\cdots$
$\displaystyle=$
$\displaystyle\frac{a_{0}}{1-t}+\frac{a_{1}zt}{1-t}+\frac{a_{2}z^{2}t^{2}}{1-t}+\cdots$
$\displaystyle=$
$\displaystyle\frac{1}{1-t}(a_{0}+a_{1}zt+a_{2}z^{2}t^{2}+\cdots)$
$\displaystyle=$ $\displaystyle\frac{1}{1-t}f(zt).$

So we obtain

$\sum_{m=0}^{\infty}T_{m}(z)t^{m}=\frac{f(zt)}{1-t}.$

Let us choose $t_{0}$ such that $|t_{0}/z|<r$. If we substitute $t_{0}/z$ for
$t$, we obtain

$\frac{f(t_{0})}{1-t_{0}/z}=\sum_{m=0}^{\infty}T_{m}(z)\left(\frac{t_{0}}{z}\right)^{m}.$

Let $z_{0}=1/z$. Then it yields that

$\frac{f(t_{0})}{1-t_{0}z_{0}}=\sum_{m=0}^{\infty}T_{m}(1/z_{0})z_{0}^{m}t_{0}^{m}.$

Observe that

$\displaystyle T_{m}(1/z_{0})z_{0}^{m}=$
$\displaystyle\left(a_{0}+a_{1}\frac{1}{z_{0}}+a_{2}\frac{1}{z_{0}^{2}}+\cdots+a_{m}\frac{1}{z_{0}^{m}}\right)z_{0}^{m}$
$\displaystyle=$ $\displaystyle
a_{0}z_{0}^{m}+a_{1}z_{0}^{m-1}+\cdots+a_{m-1}z_{0}+a_{m}$

which is the reciprocal polynomial of $T_{m}(z_{0})$. Thus

$\frac{f(t_{0})}{1-t_{0}z_{0}}$

is the generating function of the reciprocal polynomial of $T_{m}(z)$. ∎

When we work on our main theorems, we count number of zeros inside a circle
and compare this number with generated polynomial. The following theorem
becomes the tool for it.

######  Theorem 15.

[Rouché’s Theorem](Theorem 3, p177, [1]) Suppose $f$ and $g$ are analytic on
an open set containing a piece-wise smooth simple closed curve $\gamma$ and
its interior. If

$|f(z)+g(z)|<|f(z)|\text{, for all }z\in\gamma,$

then $f$ and $g$ have the same number of zeros inside $\gamma$ counting
multiplicities.

We will study some examples that follow from this theorem. These examples are
from [1].

######  Example 16.

We want to show that all the zeros of

$p(z)=3z^{3}-2z^{2}+2iz-8$

lie in the annulus $1<|z|<2$.

Let

$\displaystyle f(z)$ $\displaystyle=-3z^{3}+8$ $\displaystyle g(z)$
$\displaystyle=3z^{3}-2z^{2}+2iz-8.$

Observe that

(0.3) $|f(z)+g(z)|=|-2z^{2}+2iz|\leq 2|z|^{2}+2|z|$

and that

(0.4) $|f(z)|=|-3z^{3}+8|\geq 8-3|z|^{3}.$

On the circle $|z|=1$, in view of Inequality (0.3),

$\displaystyle|f(z)+g(z)|$ $\displaystyle\leq
2|z|^{2}+2|z|=4<5=8-3|z|^{3}\leq|f(z)|$

Meanwhile, the zeros of $f(z)=-3z^{3}+8$ are

$z=(8/3)^{1/3}e^{i2\pi/3},(8/3)^{1/3}e^{i4\pi/3},(8/3)^{1/3},$

and thus there are no zeros are inside the circle $|z|=1$. Then, given that
$f,g$ are entire, Rouché’s Theorem implies that $3z^{3}-2z^{2}+2iz-8=0$ has no
zeros inside the unit circle.

For $|z|=2$, we may deduce similarly as follows

$\displaystyle|f(z)+g(z)|$ $\displaystyle\leq 2|z|^{2}+2|z|$
$\displaystyle=8+4$ $\displaystyle<16$ $\displaystyle=3|z|^{3}-8$
$\displaystyle\leq|-3z^{3}+8|$ $\displaystyle=|f(z)|.$

Notice that all the zeros

$z=(8/3)^{1/3}e^{i2\pi/3},(8/3)^{1/3}e^{i4\pi/3},(8/3)^{1/3}$

of $f(z)$ are inside the circle $|z|=2$. Thus again, by Rouché’s Theorem,
$3z^{3}-2z^{2}+2iz-8=0$ has 3 zeros inside the circle $|z|=2$. That is, all
the zeros are inside the circle $|z|=2$. Observe that, for any
$z\in\mathbb{C}$ such that $|z|=1$,

$\displaystyle|g(z)|$ $\displaystyle=|3z^{3}-2z^{2}+2iz-8|$ $\displaystyle\geq
8-|3z^{3}-2z^{2}+2iz|$ $\displaystyle\geq 8-3|z|^{3}-2|z|^{2}-2|z|$
$\displaystyle=1.$

This implies that there is no zero on the curve $|z|=1$. All these results
combine to conclude that all the zeros lie in the annulus $1<|z|<2$.

Our next example is also from [1].

######  Example 17.

We want to find the number of the roots of the equation

$\frac{z^{2}-4}{z^{2}+4}+\frac{2z^{2}-1}{z^{2}+6}=0$

that lie within the unit circle $|z|=1$.

Observe that

$\displaystyle\frac{z^{2}-4}{z^{2}+4}+\frac{2z^{2}-1}{z^{2}+6}$
$\displaystyle=\frac{(z^{2}-4)(z^{2}+6)+(2z^{2}-1)(z^{2}+4)}{(z^{2}+4)(z^{2}+6)}$
$\displaystyle=\frac{(z^{4}+2z^{2}-24)+(2z^{4}+7z^{2}-4)}{(z^{2}+4)(z^{2}+6)}$
$\displaystyle=\frac{3z^{4}+9z^{2}-28}{(z^{2}+4)(z^{2}+6)}.$

Let

$\displaystyle f(z)$ $\displaystyle=\frac{28}{(z^{2}+4)(z^{2}+6)}$
$\displaystyle g(z)$
$\displaystyle=\frac{3z^{4}+9z^{2}-28}{(z^{2}+4)(z^{2}+6)}.$

Observe that, on $|z|=1$,

$\displaystyle|f(z)+g(z)|$
$\displaystyle=\frac{|3z^{4}+9z^{2}|}{|(z^{2}+4)(z^{2}+6)|}$
$\displaystyle\leq\frac{3|z|^{4}+9|z|^{2}}{|(z^{2}+4)(z^{2}+6)|}$
$\displaystyle=\frac{12}{|(z^{2}+4)(z^{2}+6)|}$
$\displaystyle<\frac{28}{|(z^{2}+4)(z^{2}+6)|}$
$\displaystyle=\frac{|28|}{|(z^{2}+4)(z^{2}+6)|}$ $\displaystyle=|f(z)|.$

Notice that $f,g$ are analytic on the open ball $|z|<3/2$, and that the unit
circle $|z|=1$ is inside that open ball. Then, given
$f(z)=\frac{28}{(z^{2}+4)(z^{2}+6)}$ has no zeros inside the unit circle,
Rouché’s Theorem implies that
$g(z)=\frac{3z^{4}+5z^{2}-28}{(z^{2}+4)(z^{2}+6)}$ has no zeros within the
unit circle.

In this section, we reviewed theorems and concepts from Complex Analysis. Many
theorems related to series will be used when we find explicit form of
generated sequence of functions. We review Rouché’s Theorem and studied some
examples. We will see how Rouché’s Theorem become useful when we work on our
main theorems.

## CONSEQUENCES OF THEOREMS ABOUT ZERO DISTRIBUTIONS

This paper studies the zero distribution of generated sequence of functions.
In this section we will study some special cases of studies mentioned in
‘Introduction’ for motivation. These studies are about the zero distribution
of generated sequence of functions. We first study the zero distribution of
the sequence $\\{G_{m}(z)\\}_{m\geq 0}$ generated as

$\sum_{m=0}^{\infty}G_{m}(z)t^{m}=\frac{1}{1+t+zt^{2}}.$

In particular, we prove the theorem below.

######  Theorem 18.

Let $\\{G_{m}(z)\\}$ be the sequence of polynomials generated by

$\sum_{m=0}^{\infty}G_{m}(z)t^{m}=\frac{1}{1+t+zt^{2}}.$

For any $m$, the zeros of $G_{m}(z)$ lie in the interval $(1/4,\infty)$.

This theorem is a special case of Theorem 1 in [4]. Our plan is to show that
(i) the degree of $G_{m}(z)$ is at most $\lfloor\frac{m}{2}\rfloor$ and (ii)
$G_{m}(z)$ has at least $\lfloor\frac{m}{2}\rfloor$ zeros in $(1/4,\infty)$.
Theorem 18 will then follow from the Fundamental Theorem of Algebra.

######  Lemma 19.

The sequence of polynomials $\\{G_{m}(z)\\}_{m\geq 0}$ satisfies the
recurrence

$G_{m}(z)=-G_{m-1}(z)-zG_{m-2}(z),$

for each integer $m\geq 2$. Moreover, the degree of the polynomial $G_{m}(z)$
is at most $\lfloor\frac{m}{2}\rfloor$.

###### Proof.

We first show that the sequence $\\{G_{m}(z)\\}_{m\geq 0}$ satisfies the
recurrence

$G_{m}(z)=-G_{m-1}(z)-zG_{m-2}(z)$

with the initial condition $G_{0}(z)=1$ and $G_{1}(z)=-1$. Let
$z\in\mathbb{C}$ be arbitrary. Notice that $1+t+zt^{2}$ does not have 0 as a
zero, and hence $\frac{1}{1+t+zt^{2}}$ is analytic on $B(r,0)$ for some $0<r$
where $r$ is less than both moduli of the zeros of $1+t+zt^{2}$. Theorem 6
implies the existence of a radius of convergence, say $R$, of the series
$\sum_{m=0}^{\infty}G_{m}(z)t^{m}$. Let $t\in\mathbb{C}$ such that $|t|<R$.
Then

$\sum_{m=0}^{\infty}G_{m}(z)t^{m}=\frac{1}{1+t+zt^{2}}$

holds. We multiply both sides of

$\sum_{m=0}^{\infty}G_{m}(z)t^{m}=\frac{1}{1+t+zt^{2}}$

by $1+t+zt^{2}$ to obtain

$1=(1+t+zt^{2})\sum_{m=0}^{\infty}G_{m}(z)t^{m}$

and then, by Proposition 8, given that $|t|<R$, it holds that

$\displaystyle(1+t+zt^{2})\sum_{m=0}^{\infty}G_{m}(z)t^{m}$ $\displaystyle=$
$\displaystyle
G_{0}(z)+(G_{0}(z)+G_{1}(z))t+(zG_{0}(z)+G_{1}(z)+G_{2}(z))t^{2}+\cdots.$

Consequently,

$1=G_{0}(z)+(G_{0}(z)+G_{1}(z))t+(zG_{0}(z)+G_{1}(z)+G_{2}(z))t^{2}+\cdots.$

Because of Proposition 9, we can equate the coefficients of $t^{m}$ in both
sides, and get

$\displaystyle G_{0}(z)$ $\displaystyle=1$ $\displaystyle G_{1}(z)$
$\displaystyle=-1$ $\displaystyle G_{2}(z)$
$\displaystyle=-zG_{0}(z)-G_{1}(z)$ $\displaystyle\vdots$ $\displaystyle
G_{m}(z)$ $\displaystyle=-zG_{m-2}(z)-G_{m-1}(z),m\geq 2$
$\displaystyle\vdots$

which yields the recurrence relation and the initial conditions for the
sequence $\\{G_{m}(z)\\}_{m\geq 0}$:

$\displaystyle G_{0}(z)$ $\displaystyle=1$ $\displaystyle G_{1}(z)$
$\displaystyle=-1$ $\displaystyle G_{m}(z)$
$\displaystyle=-zG_{m-2}(z)-G_{m-1}(z),m\geq 2.$

With this recurrence relation and initial condition, we can infer that
$\\{G_{m}(z)\\}_{m\geq 0}$ is a sequence of polynomials.

The next goal is prove that the degree of $G_{m}(z)$ is at most $\lfloor
m/2\rfloor$ by induction on $m$. The initial step for induction follows from
$G_{0}(z)=1$ and that $G_{1}(z)=-1$. For the induction step, we assume the
degree of $G_{k}(z)$ is at most $\left\lfloor k/2\right\rfloor$, for
$k=1,2,\dots,m-2,m-1$, for some $m\geq 2$. We consider two cases: $m$ being
even and $m$ being odd. If $m$ is even, then $m=2a$, for some
$a\in\mathbb{N}$, and consequently the degree of $G_{m-2}(z)$ is at most

$\left\lfloor\frac{m-2}{2}\right\rfloor=\left\lfloor\frac{2a-2}{2}\right\rfloor=\left\lfloor
a-1\right\rfloor=a-1,$

and the degree of $G_{m-1}(z)$ is at most

$\left\lfloor\frac{m-1}{2}\right\rfloor=\left\lfloor\frac{2a-1}{2}\right\rfloor=\left\lfloor
a-1+\frac{1}{2}\right\rfloor=a-1.$

It follows that the degree of $G_{m}(z)=-zG_{m-2}(z)-G_{m-1}(z)$ is less than
equal to the maximum of $\deg(zG_{m-2}(z))\leq a-1+1$ and
$\deg(G_{m-1}(z))\leq a-1$, and thus less than equal to

$\max(a-1,a-1+1)=a=\left\lfloor m/2\right\rfloor.$

This implies that the inductive step holds for the case $m$ is even.

If $m$ is odd, then $m=2b-1$, for some $b\in\mathbb{N}$. Observe that the
degree of $G_{m-2}(z)$ is at most

$\left\lfloor\frac{m-2}{2}\right\rfloor=\left\lfloor\frac{2b-1-2}{2}\right\rfloor=\left\lfloor
b-2+\frac{1}{2}\right\rfloor=b-2,$

and that the degree of $G_{m-1}(z)$ is at most

$\left\lfloor\frac{m-1}{2}\right\rfloor=\left\lfloor\frac{2b-1-1}{2}\right\rfloor=\left\lfloor
b-1\right\rfloor=b-1.$

It implies that the degree of $G_{m}(z)=-zG_{m-2}(z)-G_{m-1}(z)$ is less than
equal to the maximum of $\deg(zG_{m-2}(z))\leq b-2+1$ and
$\deg(G_{m-1}(z))\leq b-1$, and thus less than equal to

$\max(b-2+1,b-1)=b-1=\left\lfloor
b-1+\frac{1}{2}\right\rfloor=\left\lfloor\frac{2b-1}{2}\right\rfloor=\left\lfloor
m/2\right\rfloor.$

This concludes that the inductive step holds, and thus our claim holds. ∎

Now let us prove that $G_{m}(z)$ has at least $\lfloor\frac{m}{2}\rfloor$
zeros in $(1/4,\infty)$. The proof relies on the four functions $r(\theta)$,
$t_{1}(\theta)$, $t_{2}(\theta)$ and $z(\theta)$ defined as follows on
$\theta\in\left(\frac{\pi}{2},\pi\right)$,

(0.5) $\begin{split}r(\theta)&=-2\cos(\theta),\\\
t_{1}(\theta)&=r(\theta)e^{i\theta},\\\
t_{2}(\theta)&=r(\theta)e^{-i\theta},\\\
z(\theta)&=\frac{1}{4\cos^{2}(\theta)}.\end{split}$

######  Lemma 20.

For each $\theta\in\left(\frac{\pi}{2},\pi\right)$, the two zeros in $t$ of
$1+t+z(\theta)t^{2}$ are $t_{1}(\theta)$ and $t_{2}(\theta)$, and they are
non-zero and distinct.

###### Proof.

We will show that for each $\theta\in\left(\frac{\pi}{2},\pi\right)$

$\displaystyle t_{1}(\theta)+t_{2}(\theta)$
$\displaystyle=-\frac{1}{z(\theta)},$ $\displaystyle
t_{1}(\theta)t_{2}(\theta)$ $\displaystyle=\frac{1}{z(\theta)}.$

Indeed,

$\displaystyle t_{1}(\theta)+t_{2}(\theta)$
$\displaystyle=r(\theta)e^{i\theta}+r(\theta)e^{-i\theta}$
$\displaystyle=r(\theta)(\cos(\theta)+i\sin(\theta))+r(\theta)(\cos(\theta)-i\sin(\theta))$
$\displaystyle=2r(\theta)\cos(\theta)$ $\displaystyle=-4\cos^{2}(\theta)$
$\displaystyle=-\frac{1}{z(\theta)}.$

We also have

$t_{1}(\theta)t_{2}(\theta)=r(\theta)e^{i\theta}r(\theta)e^{-i\theta}=r(\theta)^{2}=4\cos^{2}(\theta)=\frac{1}{z(\theta)}.$

Consequently, it follows that

$\displaystyle z(\theta)(t-t_{1}(\theta))(t-t_{2}(\theta))$
$\displaystyle=z(\theta)(t^{2}-(t_{1}(\theta)+t_{2}(\theta))t+t_{1}(\theta)t_{2}(\theta))$
$\displaystyle=z(\theta)\left(t^{2}+\frac{1}{z(\theta)}t+\frac{1}{z(\theta)}\right)$
$\displaystyle=1+t+z(\theta)t^{2}$

This concludes that $t_{1}(\theta)$ and $t_{2}(\theta)$ are the zeros of
$1+t+z(\theta)t^{2}$. Notice that if $t_{1}(\theta)=t_{2}(\theta)$, then

$-\frac{1}{z(\theta)}=t_{1}(\theta)+t_{2}(\theta)=2t_{1}(\theta)$

implies that $t_{1}(\theta),t_{2}(\theta)$ are real. However, observe that,
for $\theta\in\left(\frac{\pi}{2},\pi\right)$, $r(\theta)=-2\cos(\theta)$ is
not 0 and that $e^{i\theta}$ is not real. Hence $t_{1}(\theta)=t_{2}(\theta)$
does not hold. Moreover, $t_{1}(\theta),t_{2}(\theta)\neq 0$. ∎

######  Lemma 21.

$G_{m}(z)$ has at least $\lfloor\frac{m}{2}\rfloor$ zeros on $(1/4,\infty)$.

###### Proof.

To prove this lemma, we express $G_{m}(z(\theta))$ as a function of $\theta$.
Since $t_{1}=t_{1}(\theta)$, $t_{2}=t_{2}(\theta)$ are the distinct zeros of
$1+t+z(\theta)t^{2}$, we have, for each $\theta\in(\pi/2,\pi)$,

$\frac{1}{1+t+zt^{2}}=\frac{1}{z(t-t_{1})(t-t_{2})}$

where $z=z(\theta)\neq 0$. By partial fractions, the right side is

$\frac{1}{z}\left(\frac{1}{(t-t_{1})(t_{1}-t_{2})}+\frac{1}{(t-t_{2})(t_{2}-t_{1})}\right).$

Meanwhile, observe that, for $t<\min(|t_{1}|,|t_{2}|)\neq 0$, geometric series
implies that

$\frac{1}{t-t_{1}}=\frac{1}{t_{1}}\frac{1}{\frac{t}{t_{1}}-1}=-\frac{1}{t_{1}}\frac{1}{1-\frac{t}{t_{1}}}=-\frac{1}{t_{1}}\sum_{n=0}^{\infty}\left(\frac{t}{t_{1}}\right)^{n}.$

Similarly,

$\frac{1}{t-t_{2}}=-\frac{1}{t_{2}}\sum_{n=0}^{\infty}\left(\frac{t}{t_{2}}\right)^{n}.$

We combine identities above and deduce that

$\displaystyle\sum_{m=0}^{\infty}G_{m}(z)t^{m}$ $\displaystyle=$
$\displaystyle\frac{1}{1+t+zt^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{z}\left(\left(-\frac{1}{t_{1}}\sum_{m=0}^{\infty}\left(\frac{t}{t_{1}}\right)^{m}\right)\frac{1}{(t_{1}-t_{2})}+\left(-\frac{1}{t_{2}}\sum_{m=0}^{\infty}\left(\frac{t}{t_{2}}\right)^{m}\right)\frac{1}{(t_{2}-t_{1})}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{z}\left(-\sum_{m=0}^{\infty}\frac{1}{t_{1}^{m+1}}\frac{1}{(t_{1}-t_{2})}t^{m}-\sum_{m=0}^{\infty}\frac{1}{t_{2}^{m+1}}\frac{1}{(t_{2}-t_{1})}t^{m}\right)$
$\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}-\frac{1}{z}\left(\frac{1}{t_{1}^{m+1}(t_{1}-t_{2})}+\frac{1}{t_{2}^{m+1}(t_{2}-t_{1})}\right)t^{m}.$

Thus by the uniqueness of the power series in $t$ of $(1+t+zt^{2})^{-1}$
around the origin, we conclude that

$G_{m}(z)=-\frac{1}{z}\left(\frac{1}{t_{1}^{m+1}(t_{1}-t_{2})}+\frac{1}{t_{2}^{m+1}(t_{2}-t_{1})}\right).$

We apply Definitions 0.5 to get

$\displaystyle r$ $\displaystyle=r(\theta)=-2\cos(\theta),$ $\displaystyle
t_{1}(\theta)$ $\displaystyle=r(\theta)e^{i\theta},$ $\displaystyle
t_{2}(\theta)$ $\displaystyle=r(\theta)e^{-i\theta},$ $\displaystyle
z(\theta)$
$\displaystyle=\frac{1}{r(\theta)^{2}}=\frac{1}{4\cos^{2}(\theta)},$

and conclude that, for each $\theta\in(\pi/2,\pi)$,

$\displaystyle G_{m}(z(\theta))$
$\displaystyle=-\frac{1}{z}\left(\frac{1}{t_{1}^{m+1}(t_{1}-t_{2})}+\frac{1}{t_{2}^{m+1}(t_{2}-t_{1})}\right)$
$\displaystyle=-\frac{-t_{1}^{m+1}+t_{2}^{m+1}}{zt_{1}^{m+1}t_{2}^{m+1}(t_{1}-t_{2})}$
$\displaystyle=-\frac{-r^{m+1}e^{i(m+1)\theta}+r^{m+1}e^{-i(m+1)\theta}}{\left(\frac{1}{4\cos^{2}(\theta)}\right)r^{m+1}e^{i(m+1)\theta}r^{m+1}e^{-i(m+1)\theta}(re^{i\theta}-re^{-i\theta})}$
$\displaystyle=-4\cos^{2}(\theta)\frac{-r^{m+1}e^{i(m+1)\theta}+r^{m+1}e^{-i(m+1)\theta}}{r^{2m+2}(i2r\sin(\theta))}$
$\displaystyle=4\cos^{2}(\theta)\frac{r^{m+1}(e^{i(m+1)\theta}-e^{-i(m+1)\theta})}{r^{2m+3}(i2\sin(\theta))}$
$\displaystyle=4\cos^{2}(\theta)\frac{(i2\sin((m+1)\theta)}{r^{m+2}(i2\sin(\theta))}$
$\displaystyle=\frac{4\cos^{2}(\theta)\sin((m+1)\theta)}{r^{m+2}\sin(\theta)}.$

Observe that
$\theta=\pi+\frac{1}{m+1}\pi,\pi+\frac{2}{m+1}\pi,\dots,\pi+\frac{\lfloor
m/2\rfloor}{m+1}\pi$ are the zeros of $\sin((m+1)\theta)$ and thus are the
zeros of

$G_{m}(z(\theta))=\frac{4\cos^{2}(\theta)\sin((m+1)\theta)}{r^{m+2}\sin(\theta)}.$

For $\theta=\pi+\frac{1}{m+1}\pi,\pi+\frac{2}{m+1}\pi,\dots,\pi+\frac{\lfloor
m/2\rfloor}{m+1}\pi$, notice that $z(\theta)=\frac{1}{4\cos^{2}(\theta)}$
yields $\lfloor m/2\rfloor$ number of different values. Here we have found
$\lfloor m/2\rfloor$ number of different values of $z(\theta)$ such that
$G_{m}(z(\theta))=0$. Recall that Lemma 19 says that $G_{m}(z)$ is of degree
at most $\lfloor m/2\rfloor$. The Fundamental Theorem of Algebra implies that
$G_{m}(z)$ has at most $\lfloor m/2\rfloor$ number of zeros counting
multiplicities. Since we have $\lfloor m/2\rfloor$ number of values of
$z(\theta)$ such that $G_{m}(z(\theta))=0$, those $z(\theta)$’s are all the
zeros of $G_{m}(z)=0$. Meanwhile, notice that
$z(\theta)=\frac{1}{4\cos^{2}(\theta)}$. Since $\cos^{2}(\theta)<1$, it
follows that

$z(\theta)=\frac{1}{4\cos^{2}(\theta)}>\frac{1}{4},$

for $\theta=\frac{1}{m+1}\pi,\dots,\frac{\lfloor m/2\rfloor}{m+1}\pi$. This
concludes that all the zeros of $G_{m}(z)$ lie in the interval $(1/4,\infty)$.
∎

We shall study another example, which is a spcial case of Theorem 1 in [6].

######  Theorem 22.

Let $\\{J_{m}(z)\\}_{m\geq 0}$ be the sequence of polynomials generated by

$\frac{1}{1+t+zt^{3}}.$

For each $m\geq 0$, all the zeros of $J_{m}(z)$ lie in the interval
$(-\infty,-4/27)$.

Our first goal is to prove that, for each $m\in\mathbb{N}$, $J_{m}(z)$ is of
degree at most $\lfloor m/3\rfloor$.

######  Lemma 23.

For each $m\geq 0$, $J_{m}(z)$ is a polynomial of degree at most $\lfloor
m/3\rfloor$.

###### Proof.

We first show that the sequence $\\{J_{m}(z)\\}$ satisfies the recurrence

$J_{m}(z)=-J_{m-1}(z)-zJ_{m-3}(z)$

with the initial condition

$\displaystyle J_{0}(z)$ $\displaystyle=1,$ $\displaystyle J_{1}(z)$
$\displaystyle=-1,$ $\displaystyle J_{2}(z)$ $\displaystyle=1.$

Let $z\in\mathbb{C}$ be arbitrary. Notice that $1+t+zt^{3}$ does not have 0 as
a zero, and hence $\frac{1}{1+t+zt^{3}}$ is analytic on $B(r,0)$ for some
$r>0$ where $r$ is less than all the moduli of each of the zeros of
$1+t+zt^{3}$. Then Theorem 6 implies the existence of radius of convergence,
say $R$, of the series $\sum_{m=0}^{\infty}J_{m}(z)t^{m}$. Let
$t\in\mathbb{C}$ such that $|t|<R$. Then

$\sum_{m=0}^{\infty}J_{m}(z)t^{m}=\frac{1}{1+t+zt^{3}}$

holds. We multiply both sides of

$\sum_{m=0}^{\infty}J_{m}(z)t^{m}=\frac{1}{1+t+zt^{3}}$

by $1+t+zt^{3}$ to obtain

$1=(1+t+zt^{3})\sum_{m=0}^{\infty}J_{m}(z)t^{m}$

and then, by Proposition 8, given that $|t|<R$, it holds that

$\displaystyle 1$
$\displaystyle=(1+t+zt^{3})\left(\sum_{m=0}^{\infty}J_{m}(z)t^{m}\right)$
$\displaystyle=J_{0}(z)+J_{0}(z)t+zJ_{0}(z)t^{3}+J_{1}(z)t+J_{1}(z)t^{2}+zJ_{1}(z)t^{4}+\cdots$
$\displaystyle=J_{0}(z)+\left[J_{1}(z)+J_{0}(z)\right]t+\left[J_{2}(z)+J_{1}(z)\right]t^{2}+\left[J_{3}(z)+J_{2}(z)+zJ_{0}(z)\right]t^{3}+$
$\displaystyle\cdots+\left[J_{m}(z)+J_{m-1}(z)+zJ_{m-3}(z)\right]t^{m}+\cdots$

Under Proposition 9, we may equate the coefficients of $t^{m}$ in both sides,
and it yields the following reccurence relation

$\displaystyle J_{m}(z)$ $\displaystyle=-J_{m-1}(z)-zJ_{m-3}(z)\text{, for
}m\geq 3$

with the initial condition

$\displaystyle J_{0}(z)$ $\displaystyle=1$ $\displaystyle J_{1}(z)$
$\displaystyle=-1$ $\displaystyle J_{2}(z)$ $\displaystyle=1.$

With this recurrence relation, we may deduce that $\\{J_{m}(z)\\}_{m\geq 0}$
is a sequence of polynomials.

The next step is to prove that $J_{m}(z)$ has degree at most $\lfloor
m/3\rfloor$. Observe that this claim holds for $J_{0}(z),J_{1}(z)$ and
$J_{2}(z)$. Now let us take the induction step. Suppose that the degrees of
$J_{m-3}(z),J_{m-1}(z)$ are at most $\lfloor(m-3)/3\rfloor$ and
$\lfloor(m-1)/3\rfloor$ respectively, for some $m\geq 3$. There are 3 cases:
$m\equiv 0$ (mod 3), or $m\equiv 1$ (mod 3) or $m\equiv 2$ (mod 3).

If $m\equiv 0$ (mod 3), then $m=3k$, for some $k\in\mathbb{N}$. Notice that

$\left\lfloor\frac{m-3}{3}\right\rfloor=\left\lfloor\frac{3k-3}{3}\right\rfloor=\left\lfloor
k-1\right\rfloor=k-1$

and that

$\left\lfloor\frac{m-1}{3}\right\rfloor=\left\lfloor\frac{3k-3+2}{3}\right\rfloor=\left\lfloor
k-1+\frac{2}{3}\right\rfloor=k-1.$

It follows that the degree of $J_{m}(z)=-zJ_{m-3}(z)-J_{m-1}(z)$ is at most

$\max\left(\left\lfloor\frac{m-3}{3}\right\rfloor+1,\left\lfloor\frac{m-1}{3}\right\rfloor\right)=\max\left(k-1+1,k-1\right)=k=\left\lfloor\frac{3k}{3}\right\rfloor=\left\lfloor\frac{m}{3}\right\rfloor.$

If $m\equiv 1$ (mod 3), then $m=3k+1$, for some $k\in\mathbb{N}$. Notice that

$\left\lfloor\frac{m-3}{3}\right\rfloor=\left\lfloor\frac{3k+1-3}{3}\right\rfloor=\left\lfloor
k-1+\frac{1}{3}\right\rfloor=k-1$

and that

$\left\lfloor\frac{m-1}{3}\right\rfloor=\left\lfloor\frac{3k+1-1}{3}\right\rfloor=\left\lfloor
k\right\rfloor=k.$

Then the degree of $J_{m}(z)=-zJ_{m-3}(z)-J_{m-1}(z)$ is at most

$\max\left(\left\lfloor\frac{m-3}{3}\right\rfloor+1,\left\lfloor\frac{m-1}{3}\right\rfloor\right)=k=\left\lfloor
k+\frac{1}{3}\right\rfloor=\left\lfloor\frac{3k+1}{3}\right\rfloor=\left\lfloor\frac{m}{3}\right\rfloor.$

If $m\equiv 2$ (mod 3), then $m=3k+2$, for some $k\in\mathbb{N}$. Notice that

$\left\lfloor\frac{m-3}{3}\right\rfloor=\left\lfloor\frac{3k+2-3}{3}\right\rfloor=\left\lfloor
k-1+\frac{2}{3}\right\rfloor=k-1$

and that

$\left\lfloor\frac{m-1}{3}\right\rfloor=\left\lfloor\frac{3k+2-1}{3}\right\rfloor=\left\lfloor
k+\frac{1}{3}\right\rfloor=k,$

Then the degree of $J_{m}(z)=-zJ_{m-3}(z)-J_{m-1}(z)$ is at most

$\max\left(\left\lfloor\frac{m-3}{3}\right\rfloor+1,\left\lfloor\frac{m-1}{3}\right\rfloor\right)=k=\left\lfloor
k+\frac{2}{3}\right\rfloor=\left\lfloor\frac{3k+2}{3}\right\rfloor=\left\lfloor\frac{m}{3}\right\rfloor.$

This induces that, for all cases, $J_{m}(z)$ is of degree at most
$\left\lfloor\frac{m}{3}\right\rfloor$, which implies that the inductive step
hold. It concludes that our claim holds. ∎

######  Lemma 24.

Let

(0.6) $\displaystyle r=r(\theta)$
$\displaystyle=\frac{1-4\cos^{2}(\theta)}{2\cos(\theta)},$ $\displaystyle
t_{1}$ $\displaystyle=t_{1}(\theta)=r(\theta)e^{i\theta},$ (0.7)
$\displaystyle q=q(\theta)$ $\displaystyle=-2\cos(\theta),$ $\displaystyle
t_{2}$ $\displaystyle=t_{2}(\theta)=r(\theta)e^{-i\theta},$ (0.8)
$\displaystyle z=z(\theta)$ $\displaystyle=-\frac{1}{r(\theta)^{3}q(\theta)},$
$\displaystyle t_{3}$ $\displaystyle=t_{3}(\theta)=r(\theta)q(\theta),$

where $\theta\in(2\pi/3,\pi)$. Then for each $\theta\in(2\pi/3,\pi)$,
$t_{1}(\theta),t_{2}(\theta),t_{3}(\theta)$ are distinct and non-zero. And,
for all $t\in\mathbb{C}$, it holds that

$1+t+zt^{3}=z(t-t_{1})(t-t_{2})(t-t_{3}).$

###### Proof.

Notice that

$z(t-t_{1})(t-t_{2})(t-t_{3})=z(t^{3}-(t_{1}+t_{2}+t_{3})t^{2}+(t_{1}t_{2}+t_{1}t_{3}+t_{2}t_{3})t-t_{1}t_{2}t_{3}).$

We observe that

$\displaystyle t_{1}+t_{2}+t_{3}$
$\displaystyle=r(\theta)e^{i\theta}+r(\theta)e^{-i\theta}+r(\theta)q(\theta)$
$\displaystyle=r(\theta)\left(\cos(\theta)+i\sin(\theta)+\cos(\theta)-i\sin(\theta)-2\cos(\theta)\right)$
$\displaystyle=r(\theta)(2\cos(\theta)-2\cos(\theta))$ $\displaystyle=0,$

and that

$\displaystyle t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1}=$ $\displaystyle
re^{i\theta}re^{-i\theta}+re^{-i\theta}r(-2\cos(\theta))+r(-2\cos(\theta))re^{i\theta}$
$\displaystyle=$ $\displaystyle
r(\theta)^{2}+(-2\cos(\theta))r(\theta)^{2}(e^{-i\theta}+e^{i\theta})$
$\displaystyle=$ $\displaystyle
r(\theta)^{2}+(-2\cos(\theta))r(\theta)^{2}(2\cos(\theta))$ $\displaystyle=$
$\displaystyle r(\theta)^{2}-4\cos^{2}(\theta)r(\theta)^{2}$ $\displaystyle=$
$\displaystyle r(\theta)^{2}(1-4\cos^{2}(\theta))$ $\displaystyle=$
$\displaystyle
r(\theta)^{2}\frac{1-4\cos^{2}(\theta)}{2\cos(\theta)}2\cos(\theta)$
$\displaystyle=$ $\displaystyle r(\theta)^{3}(2\cos(\theta))$ $\displaystyle=$
$\displaystyle r(\theta)^{3}(-q(\theta))$ $\displaystyle=$
$\displaystyle\frac{1}{z(\theta)}.$

Notice that

$t_{1}t_{2}t_{3}=r(\theta)e^{i\theta}r(\theta)e^{-i\theta}r(\theta)q(\theta)=r(\theta)^{3}q(\theta)=-\frac{1}{z(\theta)}.$

Above deductions imply that

$\displaystyle z(t-t_{1})(t-t_{2})(t-t_{3})$
$\displaystyle=z(t^{3}-(t_{1}+t_{2}+t_{3})t^{2}+(t_{1}t_{2}+t_{1}t_{3}+t_{2}t_{3})t-t_{1}t_{2}t_{3})$
$\displaystyle=z(\theta)\left(t^{3}-0t^{2}+\frac{1}{z(\theta)}t-\left(-\frac{1}{z(\theta)}\right)\right)$
$\displaystyle=1+t+z(\theta)t^{3}.$

Let $\theta\in(2\pi/3,\pi)$ be arbitrary. It follows that
$\cos(\theta)\in\left(-1,-\frac{1}{2}\right)$, and this induces that
$r=r(\theta)=\frac{1-4\cos^{2}(\theta)}{2\cos(\theta)}\neq 0$. Since
$\sin\theta\neq 0$, $e^{i\theta}=\cos\theta+i\sin\theta$,
$e^{-i\theta}=\cos\theta-i\sin\theta$ are distinct and not real. With
$q(\theta)=-2\cos(\theta)$ being real and non-zero, we may infer that
$q(\theta),e^{i\theta},e^{-i\theta}$ are distinct. Consequently,
$t_{1}(\theta),t_{2}(\theta),t_{3}(\theta)$ are distinct and non-zero. ∎

The next lemma gives an explicit formula for $J_{m}(z(\theta))$.

######  Lemma 25.

With $z(\theta)$, $q(\theta)$, $r(\theta)$, $t_{k}(\theta)$, $1\leq k\leq 3$,
defined as in Lemma 24, for each $\theta\in(2\pi/3,\pi)$ we have

$\displaystyle
J_{m}(z(\theta))=\frac{q^{m+1}[(-2\cos\theta)\sin(m+1)\theta-\sin(m+2)\theta]+\sin\theta}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}2ir^{2m+3}(\theta).$

###### Proof.

Let $\theta\in(2\pi/3,\pi)$ be arbitrary. We may deduce that $q(\theta)$,
$r(\theta)\neq 0$, and consequently
$t_{1}(\theta),t_{2}(\theta),t_{3}(\theta)\neq 0$. Using partial fractions
yields that, for $|t|<\min(|t_{1}|,|t_{2}|,|t_{3}|)$,

$\displaystyle\frac{1}{1+t+zt^{3}}=$
$\displaystyle\frac{1}{z(t-t_{1})(t-t_{2})(t-t_{3})}$ (0.9)
$\displaystyle\begin{split}=&\frac{1}{z}\cdot\frac{1}{(t-t_{1})(t_{2}-t_{1})(t_{3}-t_{1})}+\frac{1}{z}\cdot\frac{1}{(t-t_{2})(t_{1}-t_{2})(t_{3}-t_{2})}\\\
&+\frac{1}{z}\cdot\frac{1}{(t-t_{3})(t_{1}-t_{3})(t_{2}-t_{3})}.\end{split}$

Since $|t|<\min(|t_{1}|,|t_{2}|,|t_{3}|)$, we may deduce, with geometric
series, that expression (0.9) is equal to

$\displaystyle-\sum_{m=0}^{\infty}\frac{1}{z(t_{2}-t_{1})(t_{3}-t_{1})t_{1}^{m+1}}t^{m}-\sum_{m=0}^{\infty}\frac{1}{z(t_{1}-t_{2})(t_{3}-t_{2})t_{2}^{m+1}}t^{m}$
$\displaystyle-\sum_{m=0}^{\infty}\frac{1}{z(t_{1}-t_{3})(t_{2}-t_{3})t_{3}^{m+1}}t^{m}$

If we combine above series into one series, the coefficient of $t^{m}$ in the
series is

$\displaystyle-\frac{1}{z(t_{2}-t_{1})(t_{3}-t_{1})t_{1}^{m+1}}-\frac{1}{z(t_{1}-t_{2})(t_{3}-t_{2})t_{2}^{m+1}}-\frac{1}{z(t_{1}-t_{3})(t_{2}-t_{3})t_{3}^{m+1}}$
$\displaystyle=$
$\displaystyle-\frac{-t_{2}^{m+1}t_{3}^{m+1}(t_{2}-t_{3})-t_{1}^{m+1}t_{3}^{m+1}(t_{3}-t_{1})-t_{1}^{m+1}t_{2}^{m+1}(t_{1}-t_{2})}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}$
$\displaystyle=$
$\displaystyle\frac{t_{2}^{m+2}t_{3}^{m+1}-t_{2}^{m+1}t_{3}^{m+2}+t_{1}^{m+1}t_{3}^{m+2}-t_{1}^{m+2}t_{3}^{m+1}+t_{1}^{m+2}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{m+2}}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}$
$\displaystyle=$
$\displaystyle\frac{t_{3}^{m+2}(t_{1}^{m+1}-t_{2}^{m+1})+t_{3}^{m+1}(t_{2}^{m+2}-t_{1}^{m+2})+t_{1}^{m+2}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{m+2}}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}.$

Recall that from Equations (0.6), (0.7), (0.8) that

$\displaystyle r=r(\theta)$
$\displaystyle=\frac{1-4\cos^{2}(\theta)}{2\cos(\theta)},$ $\displaystyle
t_{1}$ $\displaystyle=t_{1}(\theta)=r(\theta)e^{i\theta},$ $\displaystyle
q=q(\theta)$ $\displaystyle=-2\cos(\theta),$ $\displaystyle t_{2}$
$\displaystyle=t_{2}(\theta)=r(\theta)e^{-i\theta},$ $\displaystyle
z=z(\theta)$ $\displaystyle=-\frac{1}{r(\theta)^{3}q(\theta)},$ $\displaystyle
t_{3}$ $\displaystyle=t_{3}(\theta)=r(\theta)q(\theta).$

These imply that

$\displaystyle\sum_{m=0}^{\infty}\left(\frac{t_{3}^{m+2}(t_{1}^{m+1}-t_{2}^{m+1})+t_{3}^{m+1}(t_{2}^{m+2}-t_{1}^{m+2})+t_{1}^{m+2}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{m+2}}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}\right)t^{m}$
$\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\left(\frac{q^{m+2}(2i\sin(m+1)\theta)+q^{m+1}(-2i\sin(m+2)\theta)+2i\sin\theta}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}\right)r^{2m+3}t^{m}$
$\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\left(\frac{q^{m+1}[(-2\cos\theta)\sin(m+1)\theta-\sin(m+2)\theta]+\sin\theta}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}\right)2ir^{2m+3}t^{m}.$

This induces that

$\displaystyle\sum_{m=0}^{\infty}J_{m}(z(\theta))t^{m}=$
$\displaystyle\frac{1}{1+t+z(\theta)t^{3}}$ $\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\left(\frac{q^{m+1}[(-2\cos\theta)\sin(m+1)\theta-\sin(m+2)\theta]+\sin\theta}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}\right)2ir^{2m+3}t^{m}$

which concludes that, for all $\theta\in(2\pi/3,\pi)$,

$J_{m}(z(\theta))=\frac{q^{m+1}[(-2\cos\theta)\sin(m+1)\theta-\sin(m+2)\theta]+\sin\theta}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}2ir^{2m+3}.\qed$

With all the lemmas at our disposal, we will prove Theorem 22.

###### Proof.

To proceed, we note that the numerator of

$\frac{q^{m+1}[(-2\cos\theta)\sin(m+1)\theta-\sin(m+2)\theta]+\sin\theta}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}},$

is $0$ when

$2\cos\theta\sin(m+1)\theta+\sin(m+2)\theta=\frac{\sin\theta}{q^{m+1}}.$

Since, for $\theta\in(2\pi/3,\pi)$,

$|q(\theta)|=|-2\cos\theta|>1$

and

$0<\sin\theta<1,$

we have

$\left|\frac{\sin\theta}{q^{m+1}}\right|<1,$

for all $\theta\in(2\pi/3,\pi)$.

Meanwhile, notice that

$\displaystyle 2\cos\theta\sin(m+1)\theta+\sin(m+2)\theta$ $\displaystyle=$
$\displaystyle\sin((m+1)\theta+\theta)+\sin((m+1)\theta-\theta)+\sin(m+2)\theta$
$\displaystyle=$ $\displaystyle 2\sin(m+2)\theta+\sin m\theta.$

When $\theta=\frac{4k-3}{2(m+2)}\pi\in(2\pi/3,\pi),k\in\mathbb{N}$, we have

$2\sin(m+2)\theta+\sin m\theta=2+\sin m\theta\geq 1.$

Similarly, if $\theta=\frac{4k-1}{2(m+2)}\pi,k\in\mathbb{N}$, then

$2\sin(m+2)\theta+\sin m\theta=-2+\sin m\theta\leq-1.$

It follows that, at
$\theta=\frac{4k-3}{2(m+2)}\pi\in(2\pi/3,\pi),k\in\mathbb{N}$,

$2\cos\theta\sin(m+1)\theta+\sin(m+2)\theta-\frac{\sin\theta}{q^{m+1}}>0$

and, at $\theta=\frac{4k-1}{2(m+2)}\pi\in(2\pi/3,\pi),k\in\mathbb{N}$,

$2\cos\theta\sin(m+1)\theta+\sin(m+2)\theta-\frac{\sin\theta}{q^{m+1}}<0.$

Then, for all $k\in\mathbb{N}$ with

$\left(\frac{4k-3}{2(m+2)}\pi,\frac{4k-1}{2(m+2)}\pi\right)\subset(2\pi/3,\pi)$

or

$\left(\frac{4k-1}{2(m+2)}\pi,\frac{4k+1}{2(m+2)}\pi\right)\subset(2\pi/3,\pi),$

the Intermediate Value Theorem induces that the equation in $\theta$

(0.10) $q^{m+1}[(-2\cos\theta)\sin(m+1)\theta-\sin(m+2)\theta]+\sin\theta=0$

has a root on the corresponding interval.

Our next goal is to show that Equation (0.10) has at least
$\left\lfloor\frac{m}{3}\right\rfloor$ solutions on the interval
$(2\pi/3,\pi)$. Observe that

$\frac{2}{3}\pi<\frac{4k+1}{2(m+2)}\pi\Leftrightarrow 2\cdot
2(m+2)<3\cdot(4k+1)\Leftrightarrow\frac{m}{3}+\frac{5}{12}<k.$

Notice that

$\displaystyle\frac{4k+3}{2(m+2)}\pi<\pi\Leftrightarrow
k<\frac{m}{2}+\frac{1}{4}.$

These inequalities combine to induce that

(0.11)
$k\in\left(\frac{m}{3}+\frac{5}{12},\frac{m}{2}+\frac{1}{4}\right)\Rightarrow\left(\frac{4k+1}{2(m+2)}\pi,\frac{4k+3}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

Likewise, observe that

$\frac{2}{3}\pi<\frac{4k+3}{2(m+2)}\pi\Leftrightarrow\frac{m}{3}-\frac{1}{12}<k,$

and

$\frac{4k+5}{2(m+2)}\pi<\pi\Leftrightarrow k<\frac{m}{2}-\frac{1}{4}.$

We get that

(0.12)
$k\in\left(\frac{m}{3}-\frac{1}{12},\frac{m}{2}-\frac{1}{4}\right)\Rightarrow\left(\frac{4k+3}{2(m+2)}\pi,\frac{4k+5}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

Now we will check 6 cases about $m$: $m\equiv 0,1,2,3,4,5\pmod{6}$.

  1. 1.

If $m\equiv 0\pmod{6}$, then $m=6n$, for some $n\geq 0$. We apply (0.11) to
infer that, for

$2n+\frac{5}{12}<k<3n+\frac{1}{4},$

that is, for $k=2n+1,2n+2,\dots,3n$, the intervals satisfy

$\left(\frac{4k+1}{2(m+2)}\pi,\frac{4k+3}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

On the other hand, notice that (0.12) induces that, for

$2n-\frac{1}{12}<k<3n-\frac{1}{4},$

that is, for $k=2n,2n+1,\dots,3n-1$, the intervals satisfy

$\left(\frac{4k+3}{2(m+2)}\pi,\frac{4k+5}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

To sum up, we have

$(3n-2n)+(3n-1-(2n-1))=2n=\left\lfloor
2n\right\rfloor=\left\lfloor\frac{6n}{3}\right\rfloor=\left\lfloor\frac{m}{3}\right\rfloor$

number of distinct intervals contained in $\left(\frac{2}{3}\pi,\pi\right)$.
Consequently, there are $\left\lfloor\frac{m}{3}\right\rfloor$ zeros.

  2. 2.

If $m\equiv 1\pmod{6}$, then $m=6n+1$ for some $n\geq 0$. We apply (0.11) to
infer that, for

$2n+\frac{9}{12}<k<3n+\frac{3}{4},$

that is, for $k=2n+1,2n+2,\dots,3n$, the intervals satisfy

$\left(\frac{4k+1}{2(m+2)}\pi,\frac{4k+3}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

On the other hand, notice that (0.12) induces that, for

$2n+\frac{3}{12}<k<3n+\frac{1}{4},$

that is, for $k=2n+1,2n+2,\dots,3n$, the intervals satisfy

$\left(\frac{4k+3}{2(m+2)}\pi,\frac{4k+5}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

To sum up, we have

$(3n-2n)+(3n-2n)=2n=\left\lfloor
2n+\frac{1}{3}\right\rfloor=\left\lfloor\frac{6n+1}{3}\right\rfloor=\left\lfloor\frac{m}{3}\right\rfloor$

number of distinct intervals contained in $\left(\frac{2}{3}\pi,\pi\right)$.
Consequently, there are $\left\lfloor\frac{m}{3}\right\rfloor$ zeros.

  3. 3.

If $m\equiv 2\pmod{6}$, then $m=6n+2$ for some $n\geq 0$. We apply (0.11) to
infer that, for

$2n+\frac{13}{12}<k<3n+\frac{5}{4},$

that is, for $k=2n+2,2n+3,\dots,3n+1$, the intervals satisfy

$\left(\frac{4k+1}{2(m+2)}\pi,\frac{4k+3}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

On the other hand, notice that (0.12) induces that ,for

$2n+\frac{7}{12}<k<3n+\frac{3}{4},$

that is, for $k=2n+1,2n+2,\dots,3n$, the intervals satisfy

$\left(\frac{4k+3}{2(m+2)}\pi,\frac{4k+5}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

To sum up, we have

$[3n+1-(2n+1)]+(3n-2n)=2n=\left\lfloor
2n+\frac{2}{3}\right\rfloor=\left\lfloor\frac{6n+2}{3}\right\rfloor=\left\lfloor\frac{m}{3}\right\rfloor$

number of distinct intervals contained in $\left(\frac{2}{3}\pi,\pi\right)$.
Consequently, there are $\left\lfloor\frac{m}{3}\right\rfloor$ zeros.

  4. 4.

If $m\equiv 3\pmod{6}$, then $m=6n+3$ for some $n\geq 0$. We apply (0.11) to
infer that, for

$2n+\frac{17}{12}<k<3n+\frac{7}{4},$

that is, for $k=2n+2,2n+3,\dots,3n+1$, the intervals satisfy

$\left(\frac{4k+1}{2(m+2)}\pi,\frac{4k+3}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

On the other hand, notice that (0.12) induces that, for

$2n+\frac{11}{12}<k<3n+\frac{5}{4},$

that is, for $k=2n+1,2n+2,\dots,3n+1$, the intervals satisfy

$\left(\frac{4k+3}{2(m+2)}\pi,\frac{4k+5}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

To sum up, we have

$[3n+1-(2n+1)]+[3n+1-(2n)]=2n+1=\left\lfloor
2n+1\right\rfloor=\left\lfloor\frac{6n+3}{3}\right\rfloor=\left\lfloor\frac{m}{3}\right\rfloor$

number of distinct intervals contained in $\left(\frac{2}{3}\pi,\pi\right)$.
Consequently, there are $\left\lfloor\frac{m}{3}\right\rfloor$ zeros.

  5. 5.

If $m\equiv 4\pmod{6}$, then $m=6n+4$ for some $n\geq 0$. We apply (0.11) to
infer that, for

$2n+\frac{21}{12}<k<3n+\frac{9}{4},$

that is, for $k=2n+2,2n+3,\dots,3n+2$, the intervals satisfy

$\left(\frac{4k+1}{2(m+2)}\pi,\frac{4k+3}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

On the other hand, notice that (0.12) induces that, for

$2n+\frac{15}{12}<k<3n+\frac{7}{4},$

that is, for $k=2n+2,2n+3,\dots,3n+1$, the intervals satisfy

$\left(\frac{4k+3}{2(m+2)}\pi,\frac{4k+5}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

To sum up, we have

$\displaystyle[3n+2-(2n+1)]+[3n+1-(2n+1)]=$ $\displaystyle 2n+1$
$\displaystyle=$ $\displaystyle\left\lfloor 2n+1+\frac{1}{3}\right\rfloor$
$\displaystyle=$ $\displaystyle\left\lfloor\frac{6n+4}{3}\right\rfloor$
$\displaystyle=$ $\displaystyle\left\lfloor\frac{m}{3}\right\rfloor$

number of distinct intervals contained in $\left(\frac{2}{3}\pi,\pi\right)$.
Consequently, there are $\left\lfloor\frac{m}{3}\right\rfloor$ zeros.

  6. 6.

If $m\equiv 5\pmod{6}$, then $m=6n+5$ for some $n\geq 0$. We apply (0.11) to
infer that, for

$2n+\frac{25}{12}<k<3n+\frac{11}{4},$

that is, for $k=2n+3,2n+4,\dots,3n+2$, the intervals satisfy

$\left(\frac{4k+1}{2(m+2)}\pi,\frac{4k+3}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

On the other hand, notice that (0.12) induces that, for

$2n+\frac{19}{12}<k<3n+\frac{9}{4},$

that is, for $k=2n+2,2n+3,\dots,3n+2$, the intervals satisfy

$\left(\frac{4k+3}{2(m+2)}\pi,\frac{4k+5}{2(m+2)}\pi\right)\subset\left(\frac{2}{3}\pi,\pi\right).$

To sum up, we have

$\displaystyle[3n+2-(2n+2)]+[3n+2-(2n+1)]=$ $\displaystyle 2n+1$
$\displaystyle=$ $\displaystyle\left\lfloor 2n+1+\frac{2}{3}\right\rfloor$
$\displaystyle=$ $\displaystyle\left\lfloor\frac{6n+5}{3}\right\rfloor$
$\displaystyle=$ $\displaystyle\left\lfloor\frac{m}{3}\right\rfloor$

number of distinct intervals contained in $\left(\frac{2}{3}\pi,\pi\right)$.
Consequently, there are $\left\lfloor\frac{m}{3}\right\rfloor$ zeros.

Here we observe that $\left(\frac{2}{3}\pi,\pi\right)$ contains at least
$\left\lfloor\frac{m}{3}\right\rfloor$ distinct zeros for all the cases
$m\equiv 0,1\dots,5\pmod{6}$.

The last thing to check is that $z(\theta)$ is injective, as a function about
$\theta$, on $\left(\frac{2}{3}\pi,\pi\right)$. Let us consider the notation
(0.8) of Lemma 24,

$z(\theta)=\frac{4\cos^{2}\theta}{(1-4\cos^{2}\theta)^{3}}.$

If we take the derivative of the function $z(\theta)$ above, we obtain

$\displaystyle z(\theta)^{\prime}$
$\displaystyle=-8\frac{\cos\theta\sin\theta}{(1-4\cos^{2}\theta)^{3}}-3\cdot
8\cos\theta\sin\theta\frac{4\cos^{2}\theta}{(1-4\cos^{2}\theta)^{4}}$
$\displaystyle=-8\cos\theta\sin\theta\left(\frac{1-4\cos^{2}\theta+12\cos^{2}\theta}{(1-4\cos^{2}\theta)^{4}}\right)$
$\displaystyle=-8\cos\theta\sin\theta\left(\frac{1+8\cos^{2}\theta}{(1-4\cos^{2}\theta)^{4}}\right).$

Notice that

$z(\theta)^{\prime}=-8\cos\theta\sin\theta\left(\frac{1+8\cos^{2}\theta}{(1-4\cos^{2}\theta)^{4}}\right)$

is positive on $\left(\frac{2}{3}\pi,\pi\right)$. Thus $z(\theta)$ is strictly
increasing on $\left(\frac{2}{3}\pi,\pi\right)$. This concludes that

$J_{m}(z(\theta))=\frac{q^{m+1}[(-2\cos\theta)\sin(m+1)\theta-\sin(m+2)\theta]+\sin\theta}{z(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})t_{1}^{m+1}t_{2}^{m+1}t_{3}^{m+1}}$

has at least $\left\lfloor\frac{m}{3}\right\rfloor$ different $z(\theta)$ on
$\left(\frac{2}{3}\pi,\pi\right)$ such that $J_{m}(z(\theta))=0$. Recall that
Lemma 23 states that $J_{m}(z)$ is of degree at most
$\left\lfloor\frac{m}{3}\right\rfloor$. Then the Fundamental Theorem of
Algebra implies that $J_{m}(z)$ has at most
$\left\lfloor\frac{m}{3}\right\rfloor$ zeros. Thus all the zeros of $J_{m}(z)$
is of the form

$z=z(\theta)=\frac{4\cos^{2}\theta}{(1-4\cos^{2}\theta)^{3}},$

where $\theta\in\left(\frac{2}{3}\pi,\pi\right)$. Recall that $z(\theta)$ is
increasing on $\left(\frac{2}{3}\pi,\pi\right)$. Thus the supremum of
$z(\theta)$ is

$z(\pi)=-\frac{4}{27}.$

This concludes that our claim holds. ∎

## POLYNOMIALS GENERATED BY $\displaystyle{\frac{1}{(at^{2}+bt+c)(1-tz)}}$

The remainder of the paper provides original results about the zeros of a
sequence of Taylor polynomials. In this section and the following sections, we
will study the pattern of zeros of sequence of functions generated as

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}$

This section is contributed to find the explicit form of generated sequence of
functions $\\{P_{m}(z)\\}_{m\geq 0}$. First we need to prove that $P_{m}(z)$
is a polynomial, for all $m\geq 0$.

######  Lemma 26.

Let $a,b,c\in\mathbb{R}$ and $c\neq 0$. Let $\\{P_{m}(z)\\}_{m\geq 0}$ be a
sequence of functions of $z$ generated as follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-zt)}.$

Then $\\{P_{m}(z)\\}_{m\geq 0}$ is a sequence of polynomials in $z$ with the
following recurrence relation

$P_{m}(z)=\frac{1}{c}\left(azP_{m-3}-(a-bz)P_{m-2}(z)-(b-cz)P_{m-1}(z)\right),$

with the initial conditions

$\displaystyle P_{0}(z)$ $\displaystyle=\frac{1}{c},$ $\displaystyle P_{1}(z)$
$\displaystyle=-\frac{1}{c^{2}}(b-cz),$ $\displaystyle P_{2}(z)$
$\displaystyle=\frac{1}{c^{3}}(b-cz)^{2}-\frac{1}{c^{2}}(a-bz).$

###### Proof.

Let $z\in\mathbb{C}$ be arbitrary. We will first find the radius of
convergence. Let us first consider the cases $a=b=z=0$. Observe that

$\frac{1}{(at^{2}+bt+c)(1-zt)}=\frac{1}{c}.$

Thus $\frac{1}{(at^{2}+bt+c)(1-zt)}$ is entire. Hence, by Theorem 6, the
radius of convergence is infinity.

For the rest of the cases, we will find an explicit $R>0$ such that
$\frac{1}{(at^{2}+bt+c)(1-zt)}$ is analytic on $B(R,0)$ and
$(at^{2}+bt+c)(1-zt)$ has a zero on $|w|=R$.

If $a=b=0$ and $z\neq 0$, observe that

$\frac{1}{(at^{2}+bt+c)(1-zt)}=\frac{1}{c(1-zt)}$

is analytic on $B(|1/z|;0)$. Notice that $1/z$ is the only zero of
$(at^{2}+bt+c)(1-zt)$. Thus $R=|1/z|$.

Now consider the cases: $a\neq 0$ or $b\neq 0$. If $a=0$ and $b\neq 0$ then

$\frac{1}{(at^{2}+bt+c)(1-zt)}=\frac{1}{(bt+c)(1-zt)}.$

which is analytic on $B(R;0)$ where

$R=\left\\{\begin{array}[]{ll}\min(\left|\frac{c}{b}\right|,|1/z|)&\text{ if
}z\neq 0\\\ \left|\frac{c}{b}\right|&\text{ if }z=0\end{array}\right.$

If $a\neq 0$, let $t_{1},t_{2}$ be the zeros of $at^{2}+bt+c$. Notice that
$\frac{1}{(at^{2}+bt+c)(1-zt)}$ is analytic on $B(R;0)$ where

$R=\left\\{\begin{array}[]{ll}\min(|t_{1}|,|t_{2}|,|1/z|)&\text{ if }z\neq
0\\\ \min(|t_{1}|,|t_{2}|)&\text{ if }z=0\end{array}\right.$

Thus in all the cases we obtain an open ball, say $B(R;0)$ in which
$\frac{1}{(at^{2}+bt+c)(1-zt)}$ is analytic and $(at^{2}+bt+c)(1-zt)$ has a
zero on $|w|=R$.

In fact, we are now going to prove that $R$ is the radius of convergence, $r$,
of the series $\sum_{m=0}^{\infty}P_{m}(z)t^{m}$. By Theorem 6, $r\geq R$.
Suppose, by contradiction, that $r>R$. By Proposition 10,
$\sum_{m=0}^{\infty}P_{m}(z)t^{m}$ is analytic as a function in $t$ on
$B(r,0)$. Since $(at^{2}+bt+c)(1-zt)$ has a root on $|w|=R$, we may infer that
$(at^{2}+bt+c)(1-zt)$ has a root on $\overline{B}(R,0)$, and let us denote the
root by $z_{0}$. Here we have $|z_{0}|=R<r$, and hence it holds that $z_{0}\in
B(r,0)$. Since $\sum_{m=0}^{\infty}P_{m}(z)t^{m}$ is analytic on $B(r,0)$, it
holds that

$\sum_{m=0}^{\infty}P_{m}(z)z_{0}^{m}=z_{1},$

for some $z_{1}\in\mathbb{C}$. However, we have that $z_{0}$ is a root of
$(at^{2}+bt+c)(1-zt)$ and thus

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-zt)}$

is not defined at $z=z_{0}$. So we encounter a contradiction. Thus we may
conclude that $R$ is the radius of convergence of
$\sum_{m=0}^{\infty}P_{m}(z)t^{m}$. This implies that

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-zt)}$

holds for $t\in B(R;0)$.

Let $t\in B(R;0)$ be arbitrary. Now let us multiply both sides by

$(at^{2}+bt+c)(1-zt)=-azt^{3}+(a-bz)t^{2}+(b-cz)t+c.$

It follows that

$(-azt^{3}+(a-bz)t^{2}+(b-cz)t+c)\sum_{m=0}^{\infty}P_{m}(z)t^{m}=1.$

If we distribute

$(-azt^{3}+(a-bz)t^{2}+(b-cz)t+c)\sum_{m=0}^{\infty}P_{m}(z)t^{m},$

we obtain

$\displaystyle-
azt^{3}\sum_{m=0}^{\infty}P_{m}(z)t^{m}+(a-bz)t^{2}\sum_{m=0}^{\infty}P_{m}(z)t^{m}+(b-cz)t\sum_{m=0}^{\infty}P_{m}(z)t^{m}$
$\displaystyle+c\sum_{m=0}^{\infty}P_{m}(z)t^{m}.$

We shall reorder the index of above series as follows:

$\displaystyle\sum_{m=3}^{\infty}-azP_{m-3}(z)t^{m}+\sum_{m=2}^{\infty}(a-bz)P_{m-2}(z)t^{m}+\sum_{m=1}^{\infty}(b-cz)P_{m-1}(z)t^{m}$
$\displaystyle+\sum_{m=0}^{\infty}cP_{m}(z)t^{m}.$

We next pull a couple of terms to make all the series start from the same
index, and we obtain

(0.13) $\begin{split}1=&\sum_{m=3}^{\infty}-azP_{m-3}(z)t^{m}\\\
&+(a-bz)P_{0}(z)t^{2}+\sum_{m=3}^{\infty}(a-bz)P_{m-2}(z)t^{m}\\\
&+(b-cz)P_{0}(z)t^{1}+(b-cz)P_{1}(z)t^{2}+\sum_{m=3}^{\infty}(b-cz)P_{m-1}(z)t^{m}\\\
&+cP_{0}(z)+cP_{1}(z)t+cP_{2}(z)t^{2}+\sum_{m=3}^{\infty}cP_{m}(z)t^{m}.\end{split}$

Since $|t|$ is less than the radius of convergence of the power series
$\sum_{m=0}^{\infty}P_{m}(z)t^{m}$, by Theorem 4, the power series
$\sum_{m=0}^{\infty}P_{m}(z)t^{m}$ converges absolutely. It follows that
$\sum_{m=0}^{\infty}|P_{m}(z)||t|^{m}$ converges.

We already fixed $t$ and we consider it as a constant. Hence

$\displaystyle|-azt^{3}|\sum_{m=0}^{\infty}|P_{m}(z)||t|^{m}$
$\displaystyle=\sum_{m=0}^{\infty}|-azP_{m}(z)t^{m+3}|,$
$\displaystyle|(a-bz)t^{2}|\sum_{m=0}^{\infty}|P_{m}(z)||t|^{m}$
$\displaystyle=\sum_{m=0}^{\infty}|(a-bz)P_{m}(z)t^{m+2}|,$
$\displaystyle|(b-cz)t|\sum_{m=0}^{\infty}|P_{m}(z)||t|^{m}$
$\displaystyle=\sum_{m=0}^{\infty}|(b-cz)P_{m}(z)t^{m+1}|,$
$\displaystyle|c|\sum_{m=0}^{\infty}|P_{m}(z)||t|^{m}$
$\displaystyle=\sum_{m=0}^{\infty}|cP_{m}(z)t^{m}|.$

Thus

$\sum_{m=3}^{\infty}-azP_{m-3}(z)t^{m},\sum_{m=2}^{\infty}(a-bz)P_{m-2}(z)t^{m},\sum_{m=1}^{\infty}(b-cz)P_{m-1}(z)t^{m},\sum_{m=0}^{\infty}cP_{m}(z)t^{m}$

are absolutely convergent on $B(R;0)$. By Proposition 10, all

$\sum_{m=3}^{\infty}-azP_{m-3}(z)t^{m},\sum_{m=2}^{\infty}(a-bz)P_{m-2}(z)t^{m},\sum_{m=1}^{\infty}(b-cz)P_{m-1}(z)t^{m},\sum_{m=0}^{\infty}cP_{m}(z)t^{m}$

are analytic. Then, by Theorem 6, all

$\sum_{m=3}^{\infty}-azP_{m-3}(z)t^{m},\sum_{m=2}^{\infty}(a-bz)P_{m-2}(z)t^{m},\sum_{m=1}^{\infty}(b-cz)P_{m-1}(z)t^{m},\sum_{m=0}^{\infty}cP_{m}(z)t^{m}$

have radius of convergence greater than or equal to $R$. Hence $R$ is less
than or equal to each of the radii of convergence of all the series:

$\sum_{m=3}^{\infty}-azP_{m-3}(z)t^{m},\sum_{m=2}^{\infty}(a-bz)P_{m-2}(z)t^{m},\sum_{m=1}^{\infty}(b-cz)P_{m-1}(z)t^{m},\sum_{m=0}^{\infty}cP_{m}(z)t^{m}$

Since we chose $t$ so that $|t|<R$, by Proposition 8, we can combine those
series as follows:

$\displaystyle\sum_{m=3}^{\infty}-azP_{m-3}(z)t^{m}+\sum_{m=3}^{\infty}(a-bz)P_{m-2}(z)t^{m}+\sum_{m=3}^{\infty}(b-cz)P_{m-1}(z)t^{m}$
$\displaystyle+\sum_{m=3}^{\infty}cP_{m}(z)t^{m}$ $\displaystyle=$
$\displaystyle\sum_{m=3}^{\infty}[-azP_{m-3}(z)+(a-bz)P_{m-2}(z)+(b-cz)P_{m-1}(z)+cP_{m}(z)]t^{m}.$

Then Equation (0.13) yields that

$\displaystyle 1=$ $\displaystyle cP_{0}(z)$
$\displaystyle+\left[cP_{1}(z)+(b-cz)P_{0}(z)\right]t$
$\displaystyle+\left[cP_{2}(z)+(b-cz)P_{1}(z)+(a-bz)P_{0}(z)\right]t^{2}$
$\displaystyle+\sum_{m=3}^{\infty}[-azP_{m-3}(z)+(a-bz)P_{m-2}(z)+(b-cz)P_{m-1}(z)+cP_{m}(z)]t^{m}.$

By Proposition 9, both sides have the same coefficient for $t^{m}$, for each
$m\geq 0$, so if we consider both sides as infinite series, we obtain that

(0.14) $\displaystyle cP_{0}(z)$ $\displaystyle=1$ (0.15) $\displaystyle
cP_{1}(z)+(b-cz)P_{0}(z)$ $\displaystyle=0$ (0.16) $\displaystyle
cP_{2}(z)+(b-cz)P_{1}(z)+(a-bz)P_{0}(z)$ $\displaystyle=0$ (0.17)
$\displaystyle-azP_{m-3}(z)+(a-bz)P_{m-2}(z)+(b-cz)P_{m-1}(z)+cP_{m}(z)$
$\displaystyle=0.$

Since we chose $z\in\mathbb{C}$ arbitrarily, this holds for all
$z\in\mathbb{C}$.

Now we will prove, by induction, that $P_{m}(z)$ is a polynomial, for all
$m\geq 0$. Let us take the base step. From Equation (0.14), we obtain

$P_{0}(z)=\frac{1}{c}.$

So $P_{0}(z)$ is a polynomial. Equation (0.15) yields that

$P_{1}(z)=-\frac{1}{c}(b-cz)P_{0}(z)=-\frac{1}{c^{2}}(b-cz)$

so $P_{1}(z)$ is a polynomial. It follows that both $(b-cz)P_{1}(z)$ and
$(a-bz)P_{0}(z)$ are polynomials. Then with Equation (0.16) we obtain

$P_{2}(z)=-\frac{1}{c}\left((b-cz)P_{1}(z)+(a-bz)P_{0}(z)\right)=\frac{1}{c^{3}}(b-cz)^{2}-\frac{1}{c^{2}}(a-bz)$

and we can observe that $-\frac{1}{c}(b-cz)P_{1}(z)-\frac{1}{c}(a-bz)P_{0}(z)$
is polynomial in $z$.

Now we will set the induction hypothesis. Assume that $P_{m-3}(z)$,
$P_{m-2}(z)$, $P_{m-1}(z)$ are polynomials. Then, by Equation (0.17), we
obtain

$P_{m}(z)=\frac{1}{c}\left(azP_{m-3}-(a-bz)P_{m-2}(z)-(b-cz)P_{m-1}(z)\right)$

which is a polynomial. This concludes that $P_{m}(z)$ is a polynomial, for all
$m=0,1,2,3,\dots$. ∎

This confirms that $P_{m}(z)$ is entire. Before we find explicit form of
$P_{m}(z)$, for the fluency of proof in some theorems, here we show an
algebraic property.

######  Proposition 27.

(0.18) $\displaystyle
t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}$
$\displaystyle=$
$\displaystyle\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)(t_{1}z-1)(t_{2}z-1).$

###### Proof.

One can check this holds for $m=0,1$. So assume $m\geq 2$. With some algebraic
calculation, we may deduce the following

(0.19) $\displaystyle
t_{1}t_{2}(t_{2}^{m}-t_{1}^{m})-t_{2}(t_{2}^{m+1}-t_{1}^{m+1})-t_{1}(t_{2}^{m+1}-t_{1}^{m+1})$
$\displaystyle=t_{1}^{m+2}-t_{2}^{m+2}$ (0.20)
$\displaystyle\begin{split}t_{1}^{k}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{k}-t_{1}^{k-1}t_{2}^{m+2}+t_{1}^{m+1}t_{2}^{k}-t_{1}^{k}t_{2}^{m+1}\\\
+t_{1}^{m+2}t_{2}^{k-1}+t_{1}^{k-1}t_{2}^{m+2}-t_{1}^{m+2}t_{2}^{k-1}&=0\end{split}$
(0.21)
$\displaystyle(t_{2}^{2}-t_{1}^{2})t_{1}^{m}t_{2}^{m}-(t_{2}-t_{1})t_{1}^{m}t_{2}^{m+1}-(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m}$
$\displaystyle=0.$

Equations (0.20) and (0.21) induce that

(0.22)
$\displaystyle\begin{split}\sum_{k=2}^{m}\left(t_{1}^{k}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{k}\right)z^{k}+\sum_{k=2}^{m}\left(-t_{1}^{k-1}t_{2}^{m+2}+t_{1}^{m+1}t_{2}^{k}\right)z^{k}&\\\
+\sum_{k=2}^{m}\left(-t_{1}^{k}t_{2}^{m+1}+t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}+\sum_{k=2}^{m}\left(t_{1}^{k-1}t_{2}^{m+2}-t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}&=0\end{split}$
(0.23)
$\displaystyle[(t_{2}^{2}-t_{1}^{2})t_{1}^{m}t_{2}^{m}-(t_{2}-t_{1})t_{1}^{m}t_{2}^{m+1}-(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m}]z^{m+1}$
$\displaystyle=0,$

respectively. If we substitute Equation (0.19) for the coefficient of $z$ in
Equation (0.18) and add Equations (0.22) and (0.23), then

$\displaystyle(t_{2}^{m+1}-t_{1}^{m+1})+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}$
(0.24)
$\displaystyle\begin{split}=&t_{2}^{m+1}-t_{1}^{m+1}+\left[t_{1}t_{2}(t_{2}^{m}-t_{1}^{m})-t_{2}(t_{2}^{m+1}-t_{1}^{m+1})-t_{1}(t_{2}^{m+1}-t_{1}^{m+1})\right]z\\\
&+\sum_{k=2}^{m}\left(t_{1}^{k}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{k}\right)z^{k}+\sum_{k=2}^{m}\left(-t_{1}^{k-1}t_{2}^{m+2}+t_{1}^{m+1}t_{2}^{k}\right)z^{k}\\\
&+\sum_{k=2}^{m}\left(-t_{1}^{k}t_{2}^{m+1}+t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}+\sum_{k=2}^{m}\left(t_{1}^{k-1}t_{2}^{m+2}-t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}\\\
&+\left[(t_{2}^{2}-t_{1}^{2})t_{1}^{m}t_{2}^{m}-(t_{2}-t_{1})t_{1}^{m}t_{2}^{m+1}-(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m}\right]z^{m+1}\\\
&+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}.\end{split}$

Observe that

  * •

${\displaystyle(t_{2}^{m+1}-t_{1}^{m+1})+(t_{1}t_{2}(t_{2}^{m}-t_{1}^{m})z+\sum_{k=2}^{m}\left(t_{1}^{k}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{k}\right)z^{k}}$
${\displaystyle=\sum_{k=0}^{m}\left(t_{1}^{k}t_{2}^{m+1}-t_{1}^{m+1}t_{2}^{k}\right)z^{k}}$

  * •

${\displaystyle-
t_{2}(t_{2}^{m+1}-t_{1}^{m+1})z-(t_{2}-t_{1})t_{1}^{m}t_{2}^{m+1}z^{m+1}+\sum_{k=2}^{m}\left(-t_{1}^{k-1}t_{2}^{m+2}+t_{1}^{m+1}t_{2}^{k}\right)z^{k}}$
${\displaystyle=\sum_{k=1}^{m+1}\left(-t_{1}^{k-1}t_{2}^{m+2}+t_{1}^{m+1}t_{2}^{k}\right)z^{k}}$

  * •

${\displaystyle-
t_{1}(t_{2}^{m+1}-t_{1}^{m+1})z-(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m}z^{m+1}+\sum_{k=2}^{m}\left(-t_{1}^{k}t_{2}^{m+1}+t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}}$
${\displaystyle=\sum_{k=1}^{m+1}\left(-t_{1}^{k}t_{2}^{m+1}+t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}}$

  * •

${\displaystyle(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}+(t_{2}^{2}-t_{1}^{2})t_{1}^{m}t_{2}^{m}z^{m+1}+\sum_{k=2}^{m}\left(t_{1}^{k-1}t_{2}^{m+2}-t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}}$
${\displaystyle=\sum_{k=2}^{m+2}\left(t_{1}^{k-1}t_{2}^{m+2}-t_{1}^{m+2}t_{2}^{k-1}\right)z^{k}.}$

Then we may infer that Expression (0.24) is equivalent to the following

$\displaystyle\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)-\left(\sum_{k=1}^{m+1}(t_{2}^{m+2-k}-t_{1}^{m+2-k})t_{1}^{k-1}t_{2}^{k}z^{k}\right)$
$\displaystyle-\left(\sum_{k=1}^{m+1}(t_{2}^{m+2-k}-t_{1}^{m+2-k})t_{1}^{k}t_{2}^{k-1}z^{k}\right)+\left(\sum_{k=2}^{m+2}(t_{2}^{m+3-k}-t_{1}^{m+3-k})t_{1}^{k-1}t_{2}^{k-1}z^{k}\right)$
$\displaystyle=$
$\displaystyle\left(\sum_{k=2}^{m+2}(t_{2}^{m+3-k}-t_{1}^{m+3-k})t_{1}^{k-1}t_{2}^{k-1}z^{k}\right)-\left(\sum_{k=1}^{m+1}(t_{2}^{m+2-k}-t_{1}^{m+2-k})t_{1}^{k}t_{2}^{k-1}z^{k}\right)$
$\displaystyle-\left(\sum_{k=1}^{m+1}(t_{2}^{m+2-k}-t_{1}^{m+2-k})t_{1}^{k-1}t_{2}^{k}z^{k}\right)+\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right).$

If we make each sums start from the index 0, above expression is equal to the
following

$\displaystyle\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k+1}t_{2}^{k+1}z^{k+2}\right)-\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k+1}t_{2}^{k}z^{k+1}\right)$
$\displaystyle-\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k+1}z^{k+1}\right)+\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right).$

We may factor out $t_{2}z$ from the first and the third sums, and then above
expression is equal to

$\displaystyle
t_{2}z\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k+1}t_{2}^{k}z^{k+1}\right)-\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k+1}t_{2}^{k}z^{k+1}\right)$
$\displaystyle-
t_{2}z\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)+\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right).$

If we factor out the common multiplicands
$\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k+1}t_{2}^{k}z^{k+1}\right)$
of the first two summands and
$\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)$
of the latter two summands, above expression is equal to

(0.25)
$\begin{split}&\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k+1}t_{2}^{k}z^{k+1}\right)(t_{2}z-1)\\\
&-\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)(t_{2}z-1).\end{split}$

Notice that

$\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k+1}t_{2}^{k}z^{k+1}=t_{1}z\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}.$

Then expression (0.25) is equal to

$\displaystyle
t_{1}z\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)(t_{2}z-1)$
$\displaystyle-\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)(t_{2}z-1).$

If we factor out the common multiplicand
$\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)(t_{2}z-1)$,
above expression is equal to

$\displaystyle\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)(t_{1}z-1)(t_{2}z-1).$

It concludes that

(0.26)
$\begin{split}&t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}\\\
=&\left(\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right)(t_{1}z-1)(t_{2}z-1).\qed\end{split}$

The following theorem shows the explicit form of $P_{m}(z)$ for
$z\in\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$.

######  Lemma 28.

Let $a,t_{1},t_{2}\in\mathbb{R}$ where $at_{1}t_{2}\neq 0$ and $t_{1}\neq
t_{2}$. Let $\\{P_{m}(z)\\}_{m\geq 0}$ be a sequence of functions of $z$
generated as follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{a(t-t_{1})(t-t_{2})(1-zt)}.$

Then, for all
$z\in\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$,

$P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}.$

###### Proof.

Let
$z\in\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$
be arbitrary. Using partial fractions, we solve for $A$, $B$, and $C$ from the
equation below

$\frac{1}{a(1-zt)(t_{1}-t)(t_{2}-t)}=\frac{A}{t_{1}-t}+\frac{B}{t_{2}-t}+\frac{C}{1-zt},$

which yields

$\displaystyle\frac{1}{a(t_{1}-t)(t_{2}-t)(1-zt)}$ $\displaystyle=$
$\displaystyle\frac{A}{t_{1}-t}+\frac{B}{t_{2}-t}+\frac{C}{1-zt}$
$\displaystyle=$
$\displaystyle\frac{A(t_{2}-t)(1-zt)+B(t_{1}-t)(1-zt)+C(t_{1}-t)(t_{2}-t)}{(t_{1}-t)(t_{2}-t)(1-zt)}$
$\displaystyle=$
$\displaystyle\frac{A[zt^{2}-(zt_{2}+1)t+t_{2}]+B[zt^{2}-(zt_{1}+1)t+t_{1}]+C[t^{2}-(t_{1}+t_{2})t+t_{1}t_{2}]}{(t_{1}-t)(t_{2}-t)(1-zt)}$
$\displaystyle=$
$\displaystyle\frac{(Az+Bz+C)t^{2}+[-A(zt_{2}+1)-B(zt_{1}+1)-(t_{1}+t_{2})C]t+At_{2}+Bt_{1}+Ct_{1}t_{2}}{(t_{1}-t)(t_{2}-t)(1-zt)}.$

This yields the following 3 equations:

(0.27) $\displaystyle Az+Bz+C$ $\displaystyle=0,$ (0.28) $\displaystyle
A(zt_{2}+1)+B(zt_{1}+1)+(t_{1}+t_{2})C$ $\displaystyle=0,$ (0.29)
$\displaystyle At_{2}+Bt_{1}+Ct_{1}t_{2}$ $\displaystyle=\frac{1}{a}.$

If we multiply Equation (0.27) by $t_{1}+t_{2}$ and subtract Equation (0.28)
from it, we get

(0.30) $A(t_{1}z-1)+B(t_{2}z-1)=0.$

If we multiply Equation (0.27) by $t_{1}t_{2}$ and subtract Equation (0.29)
from it, we get

(0.31) $A(t_{1}t_{2}z-t_{2})+B(t_{1}t_{2}z-t_{1})=-\frac{1}{a}.$

Multiplying Equation (0.30) by $t_{2}$ and subtracting Equation (0.31) from it
yields

(0.32) $B(t_{2}^{2}z-t_{2}-t_{1}t_{2}z+t_{1})=\frac{1}{a}.$

Observe that

$\displaystyle t_{2}^{2}z-t_{2}-t_{1}t_{2}z+t_{1}$
$\displaystyle=t_{2}(t_{2}z-1)-t_{1}(t_{2}z-1)$
$\displaystyle=(t_{2}-t_{1})(t_{2}z-1).$

Then we can rewrite Equation (0.32) as

$B(t_{2}-t_{1})(t_{2}z-1)=\frac{1}{a}.$

It follows that

$B=\frac{1}{a(t_{2}-t_{1})(t_{2}z-1)}.$

Also, Equation (0.30) induces

$A=-B\frac{t_{2}z-1}{t_{1}z-1}=-\frac{1}{a(t_{2}-t_{1})(t_{2}z-1)}\frac{t_{2}z-1}{t_{1}z-1}=-\frac{1}{a(t_{2}-t_{1})(t_{1}z-1)}.$

On the other hand, Equation (0.27) yields that

$\displaystyle C$ $\displaystyle=-Az-Bz$
$\displaystyle=-\left(-\frac{1}{a(t_{2}-t_{1})(t_{1}z-1)}\right)z-\left(\frac{1}{a(t_{2}-t_{1})(t_{2}z-1)}\right)z$
$\displaystyle=\frac{(t_{2}z-1)-(t_{1}z-1)}{a(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)}z$
$\displaystyle=\frac{z^{2}}{a(t_{1}z-1)(t_{2}z-1)}.$

Hence we have shown that

(0.33) $\displaystyle A=$ $\displaystyle-\frac{1}{a(t_{2}-t_{1})(t_{1}z-1)}$
(0.34) $\displaystyle B=$ $\displaystyle\frac{1}{a(t_{2}-t_{1})(t_{2}z-1)}$
(0.35) $\displaystyle C=$ $\displaystyle\frac{z^{2}}{a(t_{1}z-1)(t_{2}z-1)}.$

Meanwhile, recall that

$\displaystyle\sum_{m=0}^{\infty}P_{m}(z)t^{m}=$
$\displaystyle\frac{1}{a(t-t_{1})(t-t_{2})(1-tz)}$ $\displaystyle=$
$\displaystyle\frac{A}{t_{1}-t}+\frac{B}{t_{2}-t}+\frac{C}{1-zt}.$

Observe that, for $t<\min(|t_{1}|,|t_{2}|,|z|)$, we can apply geometric series
as follows:

$\displaystyle\frac{A}{t_{1}-t}+\frac{B}{t_{2}-t}+\frac{C}{1-zt}=$
$\displaystyle
A\frac{1}{t_{1}}\frac{1}{1-\frac{t}{t_{1}}}+B\frac{1}{t_{2}}\frac{1}{1-\frac{t}{t_{2}}}+C\frac{1}{1-zt}$
$\displaystyle=$ $\displaystyle
A\sum_{m=0}^{\infty}\frac{t^{m}}{t_{1}^{m+1}}+B\sum_{m=0}^{\infty}\frac{t^{m}}{t_{2}^{m+1}}+C\sum_{m=0}^{\infty}t^{m}z^{m}.$

Moewover, Theorem 4 implies that

$A\sum_{m=0}^{\infty}\frac{t^{m}}{t_{1}^{m+1}}+B\sum_{m=0}^{\infty}\frac{t^{m}}{t_{2}^{m+1}}+C\sum_{m=0}^{\infty}t^{m}z^{m}$

is absolutely convergent for $t\in B(\min(|t_{1}|,|t_{2}|,|z|);0)$. Then
Proposition 8 induces that

$\displaystyle
A\sum_{m=0}^{\infty}\frac{t^{m}}{t_{1}^{m+1}}+B\sum_{m=0}^{\infty}\frac{t^{m}}{t_{2}^{m+1}}+C\sum_{m=0}^{\infty}t^{m}z^{m}$
$\displaystyle=\sum_{m=0}^{\infty}t^{m}\left(\frac{A}{t_{1}^{m+1}}+\frac{B}{t_{2}^{m+1}}+Cz^{m}\right).$

Consequently, if we denote
$f(t)={\displaystyle\frac{1}{a(t-t_{1})(t-t_{2})(1-zt)}}$, by Proposition 9,
we may deduce that

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\sum_{m=0}^{\infty}\frac{1}{m!}f^{(m)}(0)t^{m}=\sum_{m=0}^{\infty}t^{m}\left(\frac{A}{t_{1}^{m+1}}+\frac{B}{t_{2}^{m+1}}+Cz^{m}\right).$

This implies that

$P_{m}(z)=\frac{A}{t_{1}^{m+1}}+\frac{B}{t_{2}^{m+1}}+Cz^{m}.$

In view of (0.33), (0.34), (0.35), we can observe that

$\displaystyle P_{m}(z)=$
$\displaystyle\frac{A}{t_{1}^{m+1}}+\frac{B}{t_{2}^{m+1}}+Cz^{m}$
$\displaystyle=$
$\displaystyle-\frac{1}{a(t_{2}-t_{1})(t_{1}z-1)}\frac{1}{t_{1}^{m+1}}$
$\displaystyle+\frac{1}{a(t_{2}-t_{1})(t_{2}z-1)}\frac{1}{t_{2}^{m+1}}$
$\displaystyle+\frac{z^{2}}{a(t_{1}z-1)(t_{2}z-1)}z^{m}$ $\displaystyle=$
$\displaystyle\frac{-(t_{2}z-1)t_{2}^{m+1}+(t_{1}z-1)t_{1}^{m+1}+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}}{a(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)t_{1}^{m+1}t_{2}^{m+1}}$
$\displaystyle=$
$\displaystyle\frac{t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}}{a(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)t_{1}^{m+1}t_{2}^{m+1}}.$

Then, by Proposition 27,

$\displaystyle P_{m}(z)=$
$\displaystyle\frac{t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}}{a(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)t_{1}^{m+1}t_{2}^{m+1}}$
$\displaystyle=$
$\displaystyle\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}.\qed$

Here we found how $P_{m}(z)$ is defined except at the points
$\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$. We can find $P_{m}(z)$
for $z=0,\frac{1}{t_{1}},\frac{1}{t_{2}}$ with the following theorem.

######  Theorem 29 (Corollary 3.8, [2]).

Let $G$ be an open connected set. If $f$ and $g$ are analytic on $G$ then
$f\equiv g$ iff $\\{z\in G;f(z)=g(z)\\}$ has a limit point in $G$.

Now we will use Theorem 29 to prove that

$P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k},$

for all $z\in\mathbb{C}$.

######  Lemma 30.

Let $a,t_{1},t_{2}\in\mathbb{R}$ where $at_{1}t_{2}\neq 0$ and $t_{1}\neq
t_{2}$. Let $\\{P_{m}(z)\\}_{m\geq 0}$ be a sequence of functions of $z$
generated as follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{a(t-t_{1})(t-t_{2})(1-zt)}.$

Then, for all $z\in\mathbb{C}$,

$P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}.$

###### Proof.

From Lemma 28, we obtain that

$P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k},$

for all
$z\in\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$.
Meanwhile, given that $at_{1}t_{2}\neq 0$, we may rewrite

$\frac{1}{a(t-t_{1})(t-t_{2})(1-zt)}=\frac{1}{(at^{2}-(t_{1}+t_{2})t+t_{1}t_{2})(1-zt)}$

where $t_{1}t_{2}\neq 0$. Then, by Lemma 26, $P_{m}(z)$ is a polynomial and
thus entire. Notice that

$\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}$

is a polynomial and thus entire. We have that

$P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k},$

for all
$z\in\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}.$
Thus $\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$
is contained in

$\left\\{z\in\mathbb{C};P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}\right\\}.$

Notice that
$\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$ has a
limit point, and thus, by Theorem 29,

$P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k},$

for all $z\in\mathbb{C}$. ∎

Now we have explicit form of $P_{m}(z)$ and we may deduce the following
corollary.

######  Corollary 31.

Suppose $ac\neq 0$ and $b^{2}-4ac\neq 0$, and $r\neq 0$. Let
$\left\\{P_{m}(z)\right\\}_{m\geq 0}$ and $\left\\{H_{m}(z)\right\\}_{m\geq
0}$ be two sequences of polynomials defined by their generating functions:

$\displaystyle\sum_{m=0}^{\infty}P_{m}(z)t^{m}$
$\displaystyle=\frac{1}{(at^{2}+bt+c)(1-tz)},$
$\displaystyle\sum_{m=0}^{\infty}H_{m}(z)t^{m}$
$\displaystyle=\frac{1}{\left[a\left(\frac{t}{r}\right)^{2}+b\left(\frac{t}{r}\right)+c\right]\left[1-tz\right]}.$

Then it holds that

$H_{m}(z)=\frac{P_{m}(rz)}{r^{m}}.$

###### Proof.

Let us denote the zeros of $at^{2}+bt+c$ by $t_{1},t_{2}$. We may deduce that
$rt_{1},rt_{2}$ are the zeros of $\frac{a}{r^{2}}t^{2}+\frac{b}{r}t+c$. Then
Lemma 30 induces that

$P_{m}(z)=\frac{1}{a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}.$

On the other hand, observe that

$\frac{1}{\left[a\left(\frac{t}{r}\right)^{2}+b\left(\frac{t}{r}\right)+c\right]\left[1-\left(\frac{t}{r}\right)\left(rz\right)\right]}=\frac{1}{\left[\frac{a}{r^{2}}t^{2}+\frac{b}{r}t+c\right]\left[1-zt\right]}$

With Lemma 30, we can infer that $H_{m}(z)$ is

$\displaystyle\frac{1}{\frac{a}{r^{2}}(rt_{2}-rt_{1})r^{m+1}t_{1}^{m+1}r^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}({t_{2}}^{m+1-k}-{t_{1}}^{m+1-k})r^{m+1-k}{(rt_{1})}^{k}{(rt_{2})}^{k}z^{k}$
$\displaystyle=$
$\displaystyle\frac{1}{r^{m}a(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}({t_{2}}^{m+1-k}-{t_{1}}^{m+1-k}){t_{1}}^{k}{t_{2}}^{k}(rz)^{k}$
$\displaystyle=$ $\displaystyle\frac{P_{m}(rz)}{r^{m}}$

This concludes that

$H_{m}(z)=\frac{P_{m}(rz)}{r^{m}}.\qed$

With this corollary, we may trace the zeros with the simpler generating
function as described in the following corollary.

######  Corollary 32.

Suppose $ac\neq 0$. Let $\left\\{P_{m}(z)\right\\}_{m\geq 0}$ and
$\left\\{H_{m}(z)\right\\}_{m\geq 0}$ be two sequences of polynomials defined
by their generating functions:

$\displaystyle\sum_{m=0}^{\infty}P_{m}(z)t^{m}$
$\displaystyle=\frac{1}{(at^{2}+bt+c)(1-tz)},$
$\displaystyle\sum_{m=0}^{\infty}H_{m}(z)t^{m}$
$\displaystyle=\frac{1}{\left(\frac{ac}{|ac|}t^{2}+\frac{b}{\sqrt{|ac|}}t+1\right)(1-tz)}.$

Then each zero of $P_{m}(z)$ is a zero of $H_{m}(z)$ multiplied by
$\left|\frac{\sqrt{|ac|}}{c}\right|$.

###### Proof.

If we consider $t_{1},t_{2}$, the zeros of $at^{2}+bt+c$, then
$at^{2}+bt+c=a(t-t_{1})(t-t_{2})$. Given that $ac\neq 0$, we obtain $c\neq 0$.
Thus we may divide both sides by $c$, and it yields the following:

$\frac{a}{c}(t-t_{1})(t-t_{2})=\frac{1}{c}(at^{2}+bt+c)=\frac{a}{c}t^{2}+\frac{b}{c}t+1.$

This implies that $\frac{a}{c}t^{2}+\frac{b}{c}t+1$ has the same zeros as
$at^{2}+bt+c$. Now let $\\{G_{m}(z)\\}_{m\geq 0}$ be a sequence of functions
of $z$ generated as follows:

$\sum_{m=0}^{\infty}G_{m}(z)t^{m}=\frac{1}{\left(\frac{a}{c}t^{2}+\frac{b}{c}t+1\right)(1-tz)}.$

By Lemma 30, we may deduce that

(0.36) $P_{m}(z)=\frac{G_{m}(z)}{c}.$

Recall that $\\{H_{m}(z)\\}_{m\geq 0}$ is a sequence of functions of $z$
generated as follows:

$\sum_{m=0}^{\infty}H_{m}(z)t^{m}=\frac{1}{\left(\frac{a}{c}\left(\frac{t}{\frac{\sqrt{|ac|}}{c}}\right)^{2}+\frac{b}{c}\frac{t}{\frac{\sqrt{|ac|}}{c}}+1\right)(1-tz)}.$

Then Corollary 31 induces that

(0.37)
$H_{m}(z)=\frac{G_{m}\left(\frac{\sqrt{|ac|}}{c}z\right)}{\left(\frac{\sqrt{|ac|}}{c}\right)^{m}},$

and Equations (0.36) and (0.37) yield that

$P_{m}(z)=\frac{G_{m}(z)}{c}=\frac{1}{c}\left(\frac{\sqrt{|ac|}}{c}\right)^{m}H_{m}\left(\frac{z}{\frac{\sqrt{|ac|}}{c}}\right)$

So, each zero of $P_{m}(z)$ is a zero of $H_{m}(z)$ multiplied by
$\frac{\sqrt{|ac|}}{c}$. ∎

## ZERO DISTRIBUTION OF $\\{P_{m}(z)\\}_{m\geq 0}$ ABOUT A CIRCLE

In this section, we will prove two of our main theorems. Let us have a look on
some examples.

Figure 0.2:  Zeros inside of the corresponding circles

Figure 0.2 shows the zeros of polynomials generated by

$\frac{1}{(t^{2}+1t-2)(1-tz)}\text{ and }\frac{1}{(t^{2}+5t+6)(1-tz)}\text{
and }\frac{1}{(t^{2}+t+2)(1-tz)}$

respectively. The left one has all the zeros inside the circle with radius
$1$. Notice that the roots of $t^{2}+1t-2=(t-1)(t+2)$ are $1$ and $-2$. The
middle figure has all the zeros outside the circle with radius $\frac{1}{2}$.
Notice that the roots of $t^{2}+5t+6=(t+2)(t+3)$ are $-2$ and $-3$. The right
figure has some zeros outside and some zeros inside the circle with radius
$\frac{1}{\sqrt{2}}$. Notice that the roots of $t^{2}+t+2=0$ are
$\frac{-1\pm\sqrt{7}i}{2}$ and the moduli are both $\sqrt{2}$.

Let us first prove a theorem about the zeros inside the circle.

######  Theorem 33.

Let $a,b,c\in\mathbb{R}$. Let $ac<0$ and $b\neq 0$. Let $\\{P_{m}(z)\\}_{m\geq
0}$ be a sequence of functions of $z$ generated as follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then all the zeros of $P_{m}(z)$ lie on the closed disk centered at 0 with
radius $1/|\alpha|$ on the complex plane $\mathbb{C}$ where $\alpha$ is the
smallest (in modulus) zero of $at^{2}+bt+c$.

###### Proof.

We will consider $\\{H_{m}(z)\\}_{m\geq 0}$, the sequence of functions of $z$
generated as follows;

$\sum_{m=0}^{\infty}H_{m}(z)t^{m}=\frac{1}{\left(-t^{2}+\frac{b}{\sqrt{-ac}}t+1\right)(1-tz)}.$

Using the Quadratic Formula, we may deduce that the zeros of
$-t^{2}+\frac{b}{\sqrt{-ac}}t+1$ are (note that $b^{2}-4ac>0$ since $ac<0$)

$\frac{-\frac{b}{\sqrt{-ac}}\pm\sqrt{-\frac{b^{2}}{ac}+4}}{-2}=\frac{1}{\sqrt{-ac}}\frac{-b\pm\sqrt{b^{2}-4ac}}{-2}.$

Let us denote those zeros by $t_{1},t_{2}$ where $|t_{1}|\leq|t_{2}|$. Let
$m\in\mathbb{N}\cup\\{0\\}$ be arbitrary.

Now we will prove that all the roots of $H_{m}(z)$ lie on the closed disk $G$
centered at 0 with radius $1/|t_{1}|$. Here the proof of Lemma 28 may be
applied to induce that

$H_{m}(z)=\frac{t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}}{-(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)t_{1}^{m+1}t_{2}^{m+1}}.$

Our goal is to use Rouché’s Theorem to show that all the zeros of the
numerator
$t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}$
are on the closed disk $|z|\leq 1/|t_{1}|$.

For the rest of the proof, we shall use the following notation:

$\displaystyle p(z)$
$\displaystyle=-(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}$ $\displaystyle
q(z)$
$\displaystyle=t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}.$

Let $\epsilon>0$ be arbitrary. Let $z$ be arbitrary such that
$|z|=1/|t_{1}|+\epsilon$. Notice that there are 2 cases: either $m$ is even or
odd. We will prove that

$|p(z)+q(z)|\leq|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon$

in both cases.

  1. 1.

$m$ is even
By Vieta’s Formula, $t_{1}t_{2}=-1$. Then one of $t_{1},t_{2}$ is negative. If
$t_{1}<0$, given $m$ is odd, then $t_{1}^{m+1}>0$. Hence it implies that

$|t_{2}^{m+1}-t_{1}^{m+1}|=t_{2}^{m+1}-t_{1}^{m+1}=|t_{2}^{m+1}|-|t_{1}^{m+1}|.$

Likewise, if $t_{2}<0$, given $m$ is even, $t_{2}^{m+1}>0$, and consequently,
we may deduce that

$|t_{2}^{m+1}-t_{1}^{m+1}|=t_{2}^{m+1}-t_{1}^{m+1}=|t_{2}^{m+1}|+|t_{1}^{m+1}|.$

In either cases $t_{1}>0$, or $t_{1}<0$, we obtain

(0.38) $|t_{2}^{m+1}-t_{1}^{m+1}|=|t_{2}^{m+1}|+|t_{1}^{m+1}|.$

Similarly, if $t_{1}<0$, then, given $m$ is even, $t_{1}^{m+2}>0$, and
consequently, we may deduce that

$|t_{1}^{m+2}-t_{2}^{m+2}|=t_{2}^{m+2}-t_{1}^{m+2}=|t_{2}^{m+2}|-|t_{1}^{m+2}|.$

Likewise, if $t_{2}<0$, given $m$ is even, $t_{2}^{m+2}>0$, and consequently,
we may deduce that

$|t_{1}^{m+2}-t_{2}^{m+2}|=t_{2}^{m+2}-t_{1}^{m+2}=|t_{2}^{m+2}|-|t_{1}^{m+2}|.$

Considering these conditions, we obtain, in either cases $t_{1}>0$, or
$t_{1}<0$, that

(0.39) $|t_{1}^{m+2}-t_{2}^{m+2}|=|t_{2}^{m+2}|-|t_{1}^{m+2}|.$

By sub-additivity,

$\displaystyle|p(z)+q(z)|=$
$\displaystyle|t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z|$
$\displaystyle\leq$
$\displaystyle|t_{2}^{m+1}-t_{1}^{m+1}|+|t_{1}^{m+2}-t_{2}^{m+2}|\cdot|z|$
$\displaystyle=$
$\displaystyle|t_{2}^{m+1}-t_{1}^{m+1}|+|t_{1}^{m+2}-t_{2}^{m+2}|\cdot\left|\frac{1}{|t_{1}|}+\epsilon\right|.$

With sub-additivity and Equations (0.39) and (0.38), we continue our deduction

$\displaystyle|t_{2}^{m+1}-t_{1}^{m+1}|+|t_{1}^{m+2}-t_{2}^{m+2}|\cdot\left|\frac{1}{|t_{1}|}+\epsilon\right|$
$\displaystyle=$
$\displaystyle|t_{2}^{m+1}|+|t_{1}^{m+1}|+(|t_{2}^{m+2}|-|t_{1}^{m+2}|)\cdot\left(\frac{1}{|t_{1}|}+\epsilon\right)$
$\displaystyle=$
$\displaystyle|t_{2}|^{m+1}+|t_{1}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}-|t_{1}|^{m+1}+(|t_{2}|^{m+2}-|t_{1}|^{m+2})\cdot\epsilon$
$\displaystyle=$
$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}-|t_{1}|^{m+2})\cdot\epsilon$
$\displaystyle\leq$
$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon.$

  2. 2.

$m$ is odd

By Vieta’s Formula, $t_{1}t_{2}=-1$. Then one of $t_{1},t_{2}$ is negative. If
$t_{1}<0$, then, given $m$ is odd, $t_{1}^{m+1}>0$, and consequently, we may
deduce that

$|t_{2}^{m+1}-t_{1}^{m+1}|=t_{2}^{m+1}-t_{1}^{m+1}=|t_{2}^{m+1}|-|t_{1}^{m+1}|.$

Likewise, if $t_{2}<0$, given $m$ is odd, $t_{2}^{m+1}>0$, and consequently,
we may deduce that

$|t_{2}^{m+1}-t_{1}^{m+1}|=t_{2}^{m+1}-t_{1}^{m+1}=|t_{2}^{m+1}|-|t_{1}^{m+1}|.$

Considering these conditions, we obtain, in either cases $t_{1}>0$, or
$t_{1}<0$, that

(0.40) $|t_{2}^{m+1}-t_{1}^{m+1}|=|t_{2}^{m+1}|-|t_{1}^{m+1}|.$

Similarly, if $t_{1}<0$, then, given $m$ is odd, $t_{1}^{m+2}<0$.
Consequently,

$|t_{1}^{m+2}-t_{2}^{m+2}|=t_{2}^{m+2}-t_{1}^{m+2}=|t_{2}^{m+2}|+|t_{1}^{m+2}|.$

Likewise, if $t_{2}<0$, given $m$ is odd, $t_{2}^{m+2}<0$. Then

$|t_{1}^{m+2}-t_{2}^{m+2}|=-t_{2}^{m+2}+t_{1}^{m+2}=|t_{2}|^{m+2}+|t_{1}|^{m+2}.$

In either cases $t_{1}>0$, or $t_{1}<0$, we obtain

(0.41) $|t_{1}^{m+2}-t_{2}^{m+2}|=|t_{2}^{m+2}|+|t_{1}^{m+2}|.$

We may deduce that, by sub-additivity,

$\displaystyle|p(z)+q(z)|=$
$\displaystyle|t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z|$
$\displaystyle\leq$
$\displaystyle|t_{2}^{m+1}-t_{1}^{m+1}|+|t_{1}^{m+2}-t_{2}^{m+2}|\cdot\left|\frac{1}{|t_{1}|}+\epsilon\right|.$

By Equations (0.40) and (0.41) and sub-additivity imply that

$\displaystyle|t_{2}^{m+1}-t_{1}^{m+1}|+|t_{1}^{m+2}-t_{2}^{m+2}|\cdot\left|\frac{1}{|t_{1}|}+\epsilon\right|$
$\displaystyle\leq$
$\displaystyle|t_{2}^{m+1}|-|t_{1}^{m+1}|+(|t_{2}^{m+2}|+|t_{1}^{m+2}|)\cdot\left(\frac{1}{|t_{1}|}+\epsilon\right)$
$\displaystyle|t_{2}^{m+1}|-|t_{1}^{m+1}|+(|t_{2}^{m+2}|+|t_{1}^{m+2}|)\cdot\left(\frac{1}{|t_{1}|}+\epsilon\right)$
$\displaystyle=$
$\displaystyle|t_{2}|^{m+1}-|t_{1}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+|t_{1}|^{m+1}+(|t_{2}|^{m+2}+|t_{1}|^{m+2})\cdot\epsilon$
$\displaystyle=$
$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}|^{m+2})\cdot\epsilon.$

Since $|t_{1}|\leq|t_{2}|$ it follows that

$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}|^{m+2})\cdot\epsilon$
$\displaystyle\leq$
$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon.$

Here we obtained, in both cases, either $m$ is even or odd, that

$|p(z)+q(z)|\leq|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon.$

Since $1\leq\left(m+2\atop{1}\right)$, it follows that

$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon$
$\displaystyle\leq$
$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon\left(m+2\atop{1}\right).$

Simplifying $|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}$ yields that

$\displaystyle|t_{2}|^{m+1}+\frac{|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon\left(m+2\atop{1}\right)$
$\displaystyle=$
$\displaystyle\frac{|t_{1}||t_{2}|^{m+1}+|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon\left(m+2\atop{1}\right).$

We may factor out $(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})$ and get

$\displaystyle\frac{|t_{1}||t_{2}|^{m+1}+|t_{2}|^{m+2}}{|t_{1}|}+(|t_{2}|^{m+2}+|t_{1}||t_{2}|^{m+1})\cdot\epsilon\left(m+2\atop{1}\right)$
$\displaystyle=$
$\displaystyle|t_{2}|^{m+1}(|t_{1}|+|t_{2}|)\left(\frac{1}{|t_{1}|}+\epsilon\left(m+2\atop{1}\right)\right).$

Multiplying and dividing by $|t_{1}|^{m+1}$, we obtain

$\displaystyle|t_{2}|^{m+1}(|t_{1}|+|t_{2}|)\left(\frac{1}{|t_{1}|}+\epsilon\left(m+2\atop{1}\right)\right)$
$\displaystyle=$
$\displaystyle|t_{1}|^{m+1}|t_{2}|^{m+1}(|t_{1}|+|t_{2}|)\left(\frac{1}{|t_{1}|^{m+2}}+\frac{1}{|t_{1}|^{m+1}}\epsilon\left(m+2\atop{1}\right)\right).$

We may infer that the last multiplicand is a part of binomial expansion
$\left(\frac{1}{|t_{1}|}+\epsilon\right)^{m+2}$. So it follows that

$\displaystyle|t_{1}|^{m+1}|t_{2}|^{m+1}(|t_{1}|+|t_{2}|)\left(\frac{1}{|t_{1}|^{m+2}}+\frac{1}{|t_{1}|^{m+1}}\epsilon\left(m+2\atop{1}\right)\right)$
$\displaystyle<$
$\displaystyle|t_{1}|^{m+1}|t_{2}|^{m+1}(|t_{1}|+|t_{2}|)\left(\frac{1}{|t_{1}|}+\epsilon\right)^{m+2}.$

Recall that $t_{1}t_{2}=-1$, thus they have opposite sign, which implies that
$|t_{1}|+|t_{2}|=|t_{1}-t_{2}|$. This continues our deduction as follows

$\displaystyle|t_{1}|^{m+1}|t_{2}|^{m+1}(|t_{1}|+|t_{2}|)\left(\frac{1}{|t_{1}|}+\epsilon\right)^{m+2}$
$\displaystyle=$
$\displaystyle|t_{1}^{m+1}t_{2}^{m+1}||t_{1}-t_{2}|\left|\frac{1}{|t_{1}|}+\epsilon\right|^{m+2}$
$\displaystyle=$
$\displaystyle\left|(t_{1}^{m+1}t_{2}^{m+1})(t_{1}-t_{2})\left(\frac{1}{|t_{1}|}+\epsilon\right)^{m+2}\right|$
$\displaystyle=$ $\displaystyle|p(z)|.$

Then, by Rouché’s Theorem (Theorem 15), all the zeros of $q(z)$ lie inside the
circle $|z|=\frac{1}{|t_{1}|}+\epsilon$. This implies that all the zeros of
$H_{m}(z)$ lie on the closed disk $|z|\leq\frac{1}{|t_{1}|}$.

Now let $z^{\prime}$ be an arbitrary root of $P_{m}(z)$. Applying Corollary
32, it follows that

$|z^{\prime}|\leq\left|\frac{\sqrt{|ac|}}{c}\frac{1}{|t_{1}|}\right|=\frac{2|a|}{-|b|+\sqrt{b^{2}-4ac}}=\frac{1}{|\alpha|}.$

This implies that all the roots of $P_{m}(z)$ lie on the closed disk centered
at 0 with radius $\frac{1}{|\alpha|}$. ∎

The next theorem propose the zeros outside of the circle

######  Theorem 34.

Let $a,b,c\in\mathbb{R}$. Suppose $ac>0$ and $b^{2}-4ac>0$. Let
$\\{P_{m}(z)\\}_{m\geq 0}$ be a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then no zeros of $P_{m}(z)$ lie inside the open ball centered at the origin
with radius $1/|\alpha|$ where $\alpha$ is the smallest (in modulus) zero of
$at^{2}+bt+c$.

###### Proof.

We will consider $\\{H_{m}(z)\\}_{m\geq 0}$, the sequence of functions of $z$
generated as follows:

$\sum_{m=0}^{\infty}H_{m}(z)t^{m}=\frac{1}{\left(t^{2}+\frac{b}{\sqrt{ac}}t+1\right)(1-tz)}.$

Notice that the zeros of $t^{2}+\frac{b}{\sqrt{ac}}t+1$ are

$\frac{-\frac{b}{\sqrt{ac}}\pm\sqrt{\frac{b^{2}}{ac}-4}}{2}=\frac{1}{\sqrt{ac}}\frac{-b\pm\sqrt{b^{2}-4ac}}{2}.$

Let us denote those zeros by $t_{1},t_{2}$ where $|t_{1}|\leq|t_{2}|$. Let
$m\in\mathbb{N}\cup\\{0\\}$ be arbitrary. Our first goal is to prove that
$H_{m}(z)$ has no root on the open ball $G$ centered at 0 with radius
$1/|t_{1}|=|t_{2}|$. Suppose, by contrary, that $H_{m}(z)$ has a root $\beta$
on the open ball $G$. Given $ac>0$ and $b^{2}-4ac>0$, it induces that
$|-b|=\sqrt{b^{2}}>\sqrt{b^{2}-4ac}$, and then

$|-b\pm\sqrt{b^{2}-4ac}|\geq|-b|-|\sqrt{b^{2}-4ac}|>0.$

This induces that $t_{1}$ and $t_{2}$ are not 0, and $b^{2}-4ac>0$ implies
that $t_{1},t_{2}$ are distinct. Lemma 30 implies that

$H_{m}(z)=\frac{1}{(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}.$

By Proposition 27,

$\displaystyle\frac{t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}}{(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)t_{1}^{m+1}t_{2}^{m+1}}$
$\displaystyle=$
$\displaystyle\frac{1}{(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k}.$

By Vieta’s Formula, we obtain $t_{1}t_{2}=1$. Then it follows that

$\displaystyle H_{m}(z)=$
$\displaystyle\frac{1}{(t_{2}-t_{1})}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})z^{k}$
$\displaystyle=$
$\displaystyle\frac{t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})z^{m+2}}{(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)}$

for
$z\in\mathbb{C}\backslash\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$.

Now let us denote

$\displaystyle p(z)=$ $\displaystyle
t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})z^{m+2},$
$\displaystyle q(z)=$
$\displaystyle-(t_{1}^{m+2}-t_{2}^{m+2})z-(t_{2}-t_{1})z^{m+2}.$

We will prove that $q(z)$ has only one zero in the open ball centered at the
origin with radius $|t_{2}|$, and then we will apply Rouché’s Theorem (c.f.
Theorem 15) for these two polynomials to find a contradiction.

Notice that

$q(z)=-(t_{1}^{m+2}-t_{2}^{m+2})z-(t_{2}-t_{1})z^{m+2}=-z[(t_{1}^{m+2}-t_{2}^{m+2})+(t_{2}-t_{1})z^{m+1}].$

Thus $z=0$ is one root of $q(z)=0$, and the other roots are the solutions of

$-(t_{1}^{m+2}-t_{2}^{m+2})-(t_{2}-t_{1})z^{m+1}=0.$

We are to prove that $|t_{1}|\neq|t_{2}|$. Suppose, by contradiction, that
$|t_{1}|=|t_{2}|$. Since $t_{1},t_{2}$ are the zeros of
$t^{2}+\frac{b}{\sqrt{ac}}t+1$, by Vieta’s Formula, $t_{1}t_{2}=1$. Then
$|t_{1}|=|t_{2}|$ implies that

$1=|t_{1}t_{2}|=|t_{1}||t_{2}|=|t_{1}|^{2}.$

This induces that $|t_{1}|=1$. We may infer that

$t_{1}=t_{2}=1\text{ or }t_{1}=t_{2}=-1.$

Notice that, by Vieta’s Formula,

$-\frac{b}{\sqrt{ac}}=t_{1}+t_{2}=\pm 2.$

Then

$\frac{b^{2}}{ac}=4,$

which yields that

$b^{2}-4ac=0.$

This contradicts against our condition $b^{2}-4ac>0$. Thus our contrary
supposition is false. So it holds that $|t_{1}|\neq|t_{2}|$. Given that
$|t_{1}|\leq|t_{2}|$, it follows that

(0.42) $|t_{1}|<|t_{2}|.$

Since $t_{1}t_{2}=1$ and $|t_{1}|<|t_{2}|$, it follows that
$|t_{1}|^{m+1}<1<|t_{2}|^{m+1}$. It induces that

$|t_{2}|^{m+2}-|t_{1}|^{m+2}>|t_{2}|^{m+2}-|t_{2}|^{m+1}|t_{1}|.$

This yields that

(0.43) $\frac{|t_{2}|^{m+2}-|t_{1}|^{m+2}}{|t_{2}|-|t_{1}|}>|t_{2}|^{m+1}.$

Meanwhile, for a zero $z_{0}$ of the equation
$-(t_{1}^{m+2}-t_{2}^{m+2})-(t_{2}-t_{1})z^{m+1}=0$, we can deduce that

(0.44)
$|z_{0}|^{m+1}=\left|\frac{t_{2}^{m+2}-t_{1}^{m+2}}{t_{2}-t_{1}}\right|.$

In view of Inequality (0.42), $|t_{1}|<|t_{2}|$, with that $t_{1}$ and $t_{2}$
have the same sign, it induces that
$|t_{2}^{m+2}-t_{1}^{m+2}|=|t_{2}|^{m+2}-|t_{1}|^{m+2}$ and that
$|t_{2}-t_{1}|=|t_{2}|-|t_{1}|$. It implies that

$\left|\frac{t_{2}^{m+2}-t_{1}^{m+2}}{t_{2}-t_{1}}\right|=\frac{|t_{2}|^{m+2}-|t_{1}|^{m+2}}{|t_{2}|-|t_{1}|}.$

Then Inequality (0.43) and Equation (0.44) yield that

$|z_{0}|^{m+1}>|t_{2}|^{m+1}.$

This yields that $|z_{0}|>|t_{2}|$, which implies that all the zeros of

$-(t_{1}^{m+2}-t_{2}^{m+2})-(t_{2}-t_{1})z^{m+1}$

are outside of the circle centered at the origin with radius $|t_{2}|$. This
concludes that

$q(z)=-(t_{1}^{m+2}-t_{2}^{m+2})z-(t_{2}-t_{1})z^{m+2}$

has only one root, $z_{0}=0$, in the open ball centered at the origin with
radius $|t_{2}|$.

Let us denote

$\displaystyle\sigma:=$
$\displaystyle(m+2)|t_{2}|^{m+1}-(|t_{2}|^{m+1}+|t_{2}|^{m}|t_{1}|+\cdots+|t_{2}||t_{1}|^{m}+|t_{1}|^{m+1}),$
$\displaystyle\zeta:=$
$\displaystyle\frac{\sigma}{\sum_{k=2}^{m+2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}}.$

Inequality (0.42) implies that $|t_{2}|^{m+2}>|t_{2}|^{m+1-k}|t_{1}|^{k}$ for
any $k=0,1,2,\dots,m+1$. This implies that $\sigma>0$. Then since
$\sum_{k=2}^{m+2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}>0$, it follows that
$\zeta>0$. Let $\epsilon>0$ be arbitrary such that

(0.45)
$\epsilon<\min\left\\{1,|t_{2}|,|t_{2}|-|t_{1}|,|t_{2}|-|\beta|,\zeta\right\\}.$

Observe that

$\displaystyle\sigma-\epsilon\sum_{k=2}^{m+2}(-\epsilon)^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}\geq\sigma-\left|\epsilon\sum_{k=2}^{m+2}(-\epsilon)^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}\right|.$

By sub-additivity,

$\displaystyle\sigma-\left|\epsilon\sum_{k=2}^{m+2}(-\epsilon)^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}\right|\geq\sigma-\epsilon\sum_{k=2}^{m+2}\epsilon^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}.$

Since $\epsilon<1$, it follows that

$\displaystyle\sigma-\epsilon\sum_{k=2}^{m+2}\epsilon^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}\geq\sigma-\epsilon\sum_{k=2}^{m+2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}.$

In view of
$\epsilon<\zeta=\frac{\sigma}{\sum_{k=2}^{m+2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}}$,
we can infer that

$\displaystyle\sigma-\epsilon\sum_{k=2}^{m+2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}$
$\displaystyle>$
$\displaystyle\sigma-\frac{\sigma}{\sum_{k=2}^{m+2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}}\sum_{k=2}^{m+2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}$
$\displaystyle=$ $\displaystyle\sigma-\sigma$ $\displaystyle=$ $\displaystyle
0.$

Consequently, we obtain

(0.46)
$\sigma-\epsilon\sum_{k=2}^{m+2}(-\epsilon)^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}>0.$

Let $\gamma$ be a circle with radius $|t_{2}|-\epsilon$ centered at $0$. Let
$z_{0}\in\gamma$ be arbitrary. Notice that

$|q(z_{0})|=|-(t_{1}^{m+2}-t_{2}^{m+2})z_{0}-(t_{2}-t_{1})z_{0}^{m+2}|.$

By sub-additivity,

$\displaystyle|-(t_{1}^{m+2}-t_{2}^{m+2})z_{0}-(t_{2}-t_{1})z_{0}^{m+2}|\geq$
$\displaystyle|-(t_{1}^{m+2}-t_{2}^{m+2})z_{0}|-|(t_{2}-t_{1})z_{0}^{m+2}|.$

In view of $|z_{0}|=|t_{2}|-\epsilon$, it follows that

$\displaystyle|-(t_{1}^{m+2}-t_{2}^{m+2})z_{0}|-|(t_{2}-t_{1})z_{0}^{m+2}|$
$\displaystyle=$
$\displaystyle(|t_{2}|^{m+2}-|t_{1}|^{m+2})(|t_{2}|-\epsilon)-(|t_{2}|-|t_{1}|)(|t_{2}|-\epsilon)^{m+2}$
$\displaystyle=$
$\displaystyle|t_{2}|^{m+3}-|t_{1}|^{m+1}-\epsilon|t_{2}|^{m+2}+\epsilon|t_{1}|^{m+2}$
$\displaystyle-(|t_{2}|-|t_{1}|)\sum_{k=0}^{m+2}(-\epsilon)^{k}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}.$

By pulling out the first two addends in the sums and then expanding those two
addends, we get

(0.47)
$\displaystyle|t_{2}|^{m+3}-|t_{1}|^{m+1}-\epsilon|t_{2}|^{m+2}+\epsilon|t_{1}|^{m+2}$
$\displaystyle-(|t_{2}|-|t_{1}|)\sum_{k=0}^{m+2}(-\epsilon)^{k}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}$
$\displaystyle=$
$\displaystyle|t_{2}|^{m+3}-|t_{1}|^{m+1}-\epsilon|t_{2}|^{m+2}+\epsilon|t_{1}|^{m+2}$
$\displaystyle-(|t_{2}|-|t_{1}|)\left(|t_{2}|^{m+2}-\epsilon(m+2)|t_{2}|^{m+1}+\sum_{k=2}^{m+2}(-\epsilon)^{k}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}\right)$
(0.48)
$\displaystyle\begin{split}=&|t_{2}|^{m+3}-|t_{1}|^{m+1}-\epsilon(|t_{2}|^{m+2}-|t_{1}|^{m+2})-|t_{2}|^{m+3}+|t_{2}|^{m+1}\\\
&+\epsilon(|t_{2}|-|t_{1}|)(m+2)|t_{2}|^{m+1}-(|t_{2}|-|t_{1}|)\sum_{k=2}^{m+2}(-\epsilon)^{k}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}.\end{split}$

After canceling $|t_{2}|^{m+3}$, we may factor as
$\epsilon(|t_{2}|^{m+2}-|t_{1}|^{m+2})=\epsilon(|t_{2}|-|t_{1}|)(|t_{2}|^{m+1}+\cdots+|t_{1}|^{m+1})$,
then we can observe that Expression (0.48) is equal to

$\displaystyle|t_{2}|^{m+1}-|t_{1}|^{m+1}-\epsilon(|t_{2}|-|t_{1}|)(|t_{2}|^{m+1}+\cdots+|t_{1}|^{m+1})$
$\displaystyle+\epsilon(|t_{2}|-|t_{1}|)(m+2)|t_{2}|^{m+1}-\epsilon^{2}(|t_{2}|-|t_{1}|)\sum_{k=2}^{m+2}(-\epsilon)^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}.$

If we factor out $\epsilon(|t_{2}|-|t_{1}|)$ from the last three addends, then
above expression is equal to

$\displaystyle|t_{2}|^{m+1}-|t_{1}|^{m+1}+\epsilon(|t_{2}|-|t_{1}|)\left(\sigma-\epsilon\sum_{k=2}^{m+2}(-\epsilon)^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}\right).$

Then, in view of Inequality (0.46), it follows that

$\displaystyle|t_{2}|^{m+1}-|t_{1}|^{m+1}+\epsilon(|t_{2}|-|t_{1}|)\left(\sigma-\epsilon\sum_{k=2}^{m+2}(-\epsilon)^{k-2}\left({m+2}\atop{k}\right)|t_{2}|^{m+2-k}\right)$
$\displaystyle>$ $\displaystyle|t_{2}|^{m+1}-|t_{1}|^{m+1}$ $\displaystyle=$
$\displaystyle|p(z_{0})+q(z_{0})|.$

By Rouché’s Theorem (c.f. Theorem 15), $p(z)$ and $q(z)$ have the same number
of zeros in the open ball centered at the origin with radius
$|t_{2}|-\epsilon$. Meanwhile, recall from condition (0.45) that

$\epsilon<\min\left\\{1,|t_{2}|,|t_{2}|-|t_{1}|,|t_{2}|-|\beta|,\zeta\right\\}.$

It induces that $\epsilon<|t_{2}|-|t_{1}|$ and $\epsilon<|t_{2}|-\beta$. It
follows that $|t_{2}|-\epsilon>|t_{1}|$, and that $|t_{2}|-\epsilon>|\beta|$.
Hence the open ball contains $t_{1}$ and $\beta$. Notice that $\beta$ is a
zero of $H_{m}(z)$ and consequently $\beta$ is a zero of $p(z)$. Meanwhile, if
we substitute $t_{1}$ for $z$ in $H_{m}(z)$, we may infer that $t_{1}$ is not
a zero of $H_{m}(z)$. But $t_{1}$ is a zero of $p(z)$. So $\beta$ and $t_{1}$
are distinct zeros of $p(z)$. Then $p(z)$ has more roots than $q(z)$ has in
the open ball, which contradicts against the deduction made by Rouché’s
Theorem. So our contrary supposition is false. This concludes that $H_{m}(z)$
has no root on the open ball centered at 0 with radius $1/|t_{1}|$.

Now let $z^{\prime}$ be arbitrary root of $P_{m}(z)$. Applying Corollary 32,
we obtain

$|z^{\prime}|\geq\frac{\sqrt{|ac|}}{|c|}\frac{1}{|t_{1}|}=\frac{2|a|}{|b|-\sqrt{b^{2}-4ac}}=\frac{1}{|\alpha|}.$

This concludes that no root of $P_{m}(z)$ lie on the open ball centered at 0
with radius $\frac{1}{|\alpha|}$. ∎

The authors of [13] introduce the history of studies about zeros of
polynomials. Gauss and Cauchy first started to study this area. The research
of this area developed throughout the history and introduced in [13]. We will
use the following theorem in this section.

######  Theorem 35 (Eneström-Kakeya’s Theorem, Theorem 1.4, [13]).

If $p(z)=\sum_{k=0}^{m}a_{k}z^{k}$ is a polynomial of degree $m$ with real and
positive coefficients, then all the zeros of $p(z)$ lie in the annulus
$r\leq|z|\leq R$ where

$r=\min_{0\leq k\leq m-1}\frac{a_{k}}{a_{k+1}}\text{ and }R=\max_{0\leq k\leq
m-1}\frac{a_{k}}{a_{k+1}}.$

We use this theorem to prove the following main theorem of this section.

######  Theorem 36.

Let $a,b,c\in\mathbb{R}\backslash\\{0\\}$. Suppose $b^{2}-4ac>0$ and $ac>0$.
Let $\\{P_{m}(z)\\}_{m\geq 0}$ be a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then no zeros of $P_{m}(z)$ lie in the closed disk centered at the origin with
radius $1/|\alpha|$, where $\alpha$ is a zero of $at^{2}+bt+c$ with greater
modulus.

###### Proof.

We will consider $\\{H_{m}(z)\\}_{m\geq 0}$, the sequence of functions of $z$
generated as follows:

$\sum_{m=0}^{\infty}H_{m}(z)t^{m}=\frac{1}{\left(\frac{ac}{|ac|}t^{2}+\frac{b}{\sqrt{|ac|}}t+1\right)(1-tz)}.$

Notice that the zeros of $\frac{ac}{|ac|}t^{2}+\frac{b}{\sqrt{|ac|}}t+1$ are

$\frac{-\frac{b}{\sqrt{|ac|}}\pm\sqrt{\frac{b^{2}}{|ac|}-4\frac{ac}{|ac|}}}{2\frac{ac}{|ac|}}=\frac{1}{\sqrt{|ac|}}\frac{-b\pm\sqrt{b^{2}-4ac}}{2\frac{ac}{|ac|}}.$

Let us denote those zeros by $t_{1},t_{2}$ where $|t_{1}|\leq|t_{2}|$.

We are to prove that $|t_{1}|\neq|t_{2}|$. Suppose, by contradiction, that
$|t_{1}|=|t_{2}|$. Since $t_{1},t_{2}$ are the zeros of
$t^{2}+\frac{b}{\sqrt{ac}}t+1$, by Vieta’s Formula, $t_{1}t_{2}=1$. Then
$|t_{1}|=|t_{2}|$ implies that

$1=|t_{1}t_{2}|=|t_{1}||t_{2}|=|t_{1}|^{2}.$

This induces that $|t_{1}|=1$. We may infer that

$t_{1}=t_{2}=1\text{ or }t_{1}=t_{2}=-1.$

Notice that, by Vieta’s Formula,

$-\frac{b}{\sqrt{ac}}=t_{1}+t_{2}=\pm 2.$

Then

$\frac{b^{2}}{ac}=4,$

which yields that

$b^{2}-4ac=0.$

This contradicts our condition $b^{2}-4ac>0$. Thus our supposition is false.
So it holds that $|t_{1}|\neq|t_{2}|$. Given that $|t_{1}|\leq|t_{2}|$, it
follows that

(0.49) $|t_{1}|<|t_{2}|.$

Recall the explicit form of $H_{m}(z)$, by Lemma 30,

$H_{m}(z)=\frac{1}{\frac{ac}{|ac|}(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})t_{1}^{k}t_{2}^{k}z^{k},m\geq
0.$

By Vieta’s Formula, $t_{1}t_{2}=\frac{|ac|}{ac}=1$. Then

$H_{m}(z)=\frac{1}{t_{2}-t_{1}}\sum_{k=0}^{m}(t_{2}^{m+1-k}-t_{1}^{m+1-k})z^{k},m\geq
0.$

Let us choose $m\geq 1$ arbitrarily. We are to prove that
$\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}>\frac{1}{|t_{1}|}$
for $k=1,2,\dots,m-1$. Let $0\leq k\leq m-1$ be arbitrary. With
$t_{1}t_{2}=1$, we may deduce that

$\displaystyle|t_{2}^{m+1-k}-t_{1}^{m+1-k}|\cdot|t_{1}|=$
$\displaystyle|t_{2}^{m+1-k}t_{1}-t_{1}^{m+2-k}|$ $\displaystyle=$
$\displaystyle|t_{2}^{m-k}-t_{1}^{m+2-k}|.$

Since $t_{1},t_{2}$ have the same sign, it can be deduced that

$\displaystyle|t_{2}^{m-k}-t_{1}^{m+2-k}|=$
$\displaystyle|t_{2}|^{m-k}-|t_{1}|^{m+2-k}.$

Inequality (0.49), in view of $t_{1}t_{2}=1$, induces that $|t_{1}|<1$. It
follows that

$\displaystyle|t_{2}|^{m-k}-|t_{1}|^{m+2-k}>|t_{2}|^{m-k}-|t_{1}|^{m-k}=|t_{2}^{m-k}-t_{1}^{m-k}|.$

Consequently we obtain

$|t_{2}^{m+1-k}-t_{1}^{m+1-k}|\cdot|t_{1}|>|t_{2}^{m-k}-t_{1}^{m-k}|,$

which yields that

(0.50)
$\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}>\frac{1}{|t_{1}|}.$

With this Inequality (0.50), we will prove that all the zeros of $H_{m}(z)$
are outside of the circle $|z|=\frac{1}{|t_{1}|}$. Denote, for each $0\leq
k\leq m$, the coefficient of $z^{k}$ by $a_{k}$ as follows

$a_{k}:=\frac{t_{2}^{m+1-k}-t_{1}^{m+1-k}}{t_{2}-t_{1}}.$

If $t_{2}>0$, then, under Inequality (0.49),

$t_{2}-t_{1}=|t_{2}|-t_{1}\geq|t_{2}|-|t_{1}|>0$

and

$t_{2}^{m+1-k}-t_{1}^{m+1-k}=|t_{2}|^{m+1-k}-t_{1}^{m+1-k}\geq|t_{2}|^{m+1-k}-|t_{1}|^{m+1-k}>0.$

Hence $a_{k}$ is positive. So $H_{m}(z)$ is a polynomial with all positive
coefficients. Thus

$\frac{a_{k}}{a_{k+1}}=\frac{|a_{k}|}{|a_{k+1}|}=\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}.$

Here Inequality (0.50) implies that

$\min_{0\leq k\leq m-1}\frac{a_{k}}{a_{k+1}}=\min_{0\leq k\leq
m-1}\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}>\frac{1}{|t_{1}|}.$

Then Theorem 35 induces that all the zeros of $H_{m}(z)$ are outside of the
circle $|z|=\frac{1}{|t_{1}|}$.

If $t_{2}<0$, since one of $m-k$ and $m+1-k$ is odd the the other is even, one
of $t_{2}^{m-k}$ and $t_{2}^{m+1-k}$ is positive and the other is negative. We
may infer that

(0.51)
$\frac{a_{k}}{a_{k+1}}=\frac{t_{2}^{m+1-k}-t_{1}^{m+1-k}}{t_{2}^{m-k}-t_{1}^{m-k}}<0.$

Then the sign of $a_{k}$ alternate, since otherwise, there would exist
$a_{k^{\prime}},a_{k^{\prime}+1}$ with the same sign, and it would yield that

$\frac{a_{k^{\prime}}}{a_{k^{\prime}+1}}>0$

which contradicts against Inequality (0.51). Notice that

$a_{0}=\frac{t_{2}^{m+1}-t_{1}^{m+1}}{t_{2}-t_{1}}.$

Notice that

(0.52) $t_{2}-t_{1}=-|t_{2}|-t_{1}\leq-|t_{2}|+|t_{1}|<0.$

If $m$ is odd, then $m+1$ is even. It follows that

$t_{2}^{m+1}-t_{1}^{m+1}=|t_{2}|^{m+1}-|t_{1}|^{m+1}>0.$

Considering Inequality (0.52), it holds that $a_{0}<0$. We may infer that

$-H_{m}(-z)$

have all positive coefficients. Denote the coefficient of $z^{k}$ of
$-H_{m}(-z)$ by $b_{k}$. Notice that

$b_{k}=\frac{t_{2}^{m+1-k}-t_{1}^{m+1-k}}{t_{2}-t_{1}}(-1)^{k+1}.$

We may deduce that

$\frac{b_{k}}{b_{k+1}}=\frac{|b_{k}|}{|b_{k+1}|}=\frac{|a_{k}|}{|a_{k+1}|}=\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}.$

Here Inequality (0.50) implies that

$\min_{0\leq k\leq m-1}\frac{b_{k}}{b_{k+1}}=\min_{0\leq k\leq
m-1}\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}>\frac{1}{|t_{1}|}.$

Then Theorem 35 induces that all the zeros of $-H_{m}(-z)$ are outside of the
circle $|z|=\frac{1}{|t_{1}|}$. We may infer that all the zeros of $H_{m}(z)$
are outside of the circle $|z|=\frac{1}{|t_{1}|}$.

If $m$ is even, then $m+1$ is odd. It follows that

$t_{2}^{m+1}-t_{1}^{m+1}=-|t_{2}|^{m+1}+|t_{1}|^{m+1}<0.$

Considering Inequality (0.52), it holds that $a_{0}>0$ We may infer that

$H_{m}(-z)$

have all positive coefficients. Denote the coefficient of $z^{k}$ of
$H_{m}(-z)$ by $c_{k}$. Notice that

$c_{k}=\frac{t_{2}^{m+1-k}-t_{1}^{m+1-k}}{t_{2}-t_{1}}(-1)^{k}.$

We may deduce that

$\frac{c_{k}}{c_{k+1}}=\frac{|c_{k}|}{|c_{k+1}|}=\frac{|a_{k}|}{|a_{k+1}|}=\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}.$

Here Inequality (0.50) implies that

$\min_{0\leq k\leq m-1}\frac{c_{k}}{c_{k+1}}=\min_{0\leq k\leq
m-1}\frac{|t_{2}^{m+1-k}-t_{1}^{m+1-k}|}{|t_{2}^{m-k}-t_{1}^{m-k}|}>\frac{1}{|t_{1}|}.$

Then Eneström- Kakeya’s Theorem 35 induces that all the zeros of $H_{m}(-z)$
are outside of the circle $|z|=\frac{1}{|t_{1}|}$. We may infer that all the
zeros of $H_{m}(z)$ are outside of the circle $|z|=\frac{1}{|t_{1}|}$.

This concluds that, in all cases, the zeros of $H_{m}(z)$ are outside of the
circle $|z|=\frac{1}{|t_{1}|}$. ∎

## APPLICATION OF LIMIT DISTRIBUTION OF ZEROS

In the previous section, we showed that the zeros of $P_{m}(z)$ are related to
the circle centered at the origin with radius $1/|\alpha|$ where $\alpha$ is
one of the roots of $at^{2}+bt+c$ with smallest modulus. In this section we
will study the ‘convergence’ of zeros.

Figure 0.3:  Zeros of $P_{m}(z)$ for $m=15,30,100$

Figure 0.3 shows the zeros of $P_{15}(z),P_{30}(z),P_{100}(z)$ generated as

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(t^{2}+t-2)(1-zt)}.$

The zeros are becoming denser and shaping a circle $|z|=1$. The main theorem
of this section proves that the zeros converges to the circle. We now
rigorously define what ‘convergence’ is. First, let us define the set of
zeros.

######  Definition 37 ((1.5), [5]).

Suppose $\\{f_{m}\\}_{m\geq 0}$ analytic in a region (an open and connected
set) $D$. We denote, for each $m\geq 0$, $\mathcal{Z}(f_{m})$ the set of zeros
of $f_{m}$ in $D$.

There are two kind of ‘limit’ of zeros.

######  Definition 38 ((1.7), [5]).

We denote $\overline{\lim}\,\mathcal{Z}(f_{m})$ the set of $z^{*}\in D$ such
that

$\forall r>0,\forall N\in\mathbb{N},\exists
m>N;B(r,z^{*})\cap\mathcal{Z}(f_{m})\neq\emptyset.$

######  Definition 39 ((1.6), [5]).

We denote $\underline{\lim}\,\mathcal{Z}(f_{m})$ to be the set of $z^{*}\in D$
such that

$\forall r>0,\exists N_{r}\in\mathbb{N};\forall
m>N_{r},B(r,z^{*})\cap\mathcal{Z}(f_{m})\neq\emptyset.$

The two limits above have the following property.

######  Proposition 40.

Suppose $\\{f_{m}\\}_{m\geq 0}$ analytic in a region $D$. Then

$\underline{\lim}\,\mathcal{Z}(f_{m})\subseteq\overline{\lim}\,\mathcal{Z}(f_{m})$

###### Proof.

Let $z^{*}\in\underline{\lim}\,\mathcal{Z}(f_{m})$ be arbitrary. Let $r>0$ and
$N\in\mathbb{N}$ be arbitrary. Under the definition of
$\underline{\lim}\,\mathcal{Z}(f_{m})$, there exists $N_{r}\in\mathbb{N}$ such
that, for any $m>N_{r}$, it holds that
$B(r,z^{*})\cap\mathcal{Z}(f_{m})\neq\emptyset$. Then, for
$n_{0}=\max(N_{r},N)$ we obtain
$B(r,z^{*})\cap\mathcal{Z}(f_{n_{0}})\neq\emptyset$. This implies that
$z^{*}\in\overline{\lim}\,\mathcal{Z}(f_{m})$. Thus,

$\underline{\lim}\,\mathcal{Z}(f_{m})\subseteq\overline{\lim}\,\mathcal{Z}(f_{m}).\qed$

When those two limits are the same we define a new limit.

######  Definition 41.

If $\underline{\lim}\,\mathcal{Z}(f_{m})=\overline{\lim}\,\mathcal{Z}(f_{m})$,
then we define

$\lim\mathcal{Z}(f_{m}):=\underline{\lim}\,\mathcal{Z}(f_{m})=\overline{\lim}\,\mathcal{Z}(f_{m}).$

The main theorem in this section will find the set $\lim\mathcal{Z}(P_{m})$.
To confirm the existence of $\lim\mathcal{Z}(P_{m})$, and to find the elements
of $\lim\mathcal{Z}(P_{m})$ the following theorem provides a convenient tool
for us.

######  Theorem 42.

[Theorem 1.5, [5]] Suppose $\\{f_{m}\\}_{m\geq 0}$ is a sequence of functions
analytic in an open connected set (region) $D$. Suppose that there exists
$n\in\mathbb{N}$ and $\alpha_{k}(z),\beta_{k}(z)$, $1\leq k\leq n$, analytic
in $D$ such that

$f_{m}(z)=\sum_{k=1}^{n}\alpha_{k}(z)\beta_{k}(z)^{m}\text{, for all
}m\in\mathbb{N},$

and there does not exist $i\neq j$ such that $\beta_{i}(z)=w\beta_{j}(z)$ for
any $z\in D$, for some $w$ with $|w|=1$. Then
$\underline{\lim}\,\mathcal{Z}(f_{m})=\overline{\lim}\,\mathcal{Z}(f_{m})$.
Furthermore, $z^{*}\in\lim\mathcal{Z}(f_{m})$ if and only if one of the
following holds

  * •

$\exists i\neq j$ such that
$|\beta_{i}(z^{*})|=|\beta_{j}(z^{*})|\geq|\beta_{k}(z^{*})|$, for all
$k=1,2,\dots,n$, and $\alpha_{i}(z^{*}),\alpha_{j}(z^{*})\neq 0$

  * •

$\exists i$ such that $|\beta_{i}(z^{*})|>|\beta_{k}(z^{*})|$, for all
$k=1,2,\dots,n$, $k\neq i$, and $\alpha_{i}(z^{*})=0$

Before get into our main theorem, let us study an example of this.

######  Example 43.

Let us define, for each $m\in\mathbb{N}$,

$f_{m}(z):=(z-1)^{m}+z^{m}+(z+1)^{m}.$

Then

$\lim\mathcal{Z}(f_{m})=\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}.$

Figure 0.4:  Zeros of $f_{m}(z)$ for $m=10,15,50$

The points in the Figure 0.4 become denser as $m$ increase.

Notice that

$f_{m}(z)=\sum_{k=1}^{3}\alpha_{k}(z)\beta_{k}(z)^{m}\text{, for all
}m\in\mathbb{N}$

where

$\displaystyle\alpha_{1}(z)=$ $\displaystyle 1$ $\displaystyle\beta_{1}(z)=$
$\displaystyle z-1$ $\displaystyle\alpha_{2}(z)=$ $\displaystyle 1$
$\displaystyle\beta_{2}(z)=$ $\displaystyle z$ $\displaystyle\alpha_{3}(z)=$
$\displaystyle 1$ $\displaystyle\beta_{3}(z)=$ $\displaystyle z+1.$

For any $w\in\mathbb{C}$ with $|w|=1$, we can observe that $|\beta_{1}(-1)|=2$
and $|w\beta_{2}(-1)|=1$. Notice that, for any $w\in\mathbb{C}$ with $|w|=1$,
we can observe that $|\beta_{2}(-1)|=1$ and $|w\beta_{3}(-1)|=0$. Likewise,
for any $w\in\mathbb{C}$ with $|w|=1$, we can observe that $|\beta_{3}(-1)|=0$
and $|w\beta_{1}(-1)|=2$. So, there does not exist $i\neq j$ such that
$\beta_{i}(z)=w\beta_{j}(z)$ for any $z\in D$, for some $w$ with $|w|=1$. This
implies, by Theorem 42, that
$\underline{\lim}\,\mathcal{Z}(f_{m})=\overline{\lim}\,\mathcal{Z}(f_{m})$.
Thus $\lim\mathcal{Z}(f_{m})$ exists.

Now let us find $\lim\mathcal{Z}(f_{m})$. Let
$z^{*}\in\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}$ be arbitrary. Then
$z^{*}=yi$ for some $y\in\mathbb{R}$. Observe that

$|\beta_{1}(z^{*})|=|yi-1|=\sqrt{y^{2}+1}=|yi+1|=|\beta_{3}(z^{*})|$

and that

$|\beta_{2}(z^{*})|=|yi|=|y|<\sqrt{y^{2}+1}=|\beta_{1}(z^{*})|=|\beta_{3}(z^{*})|.$

Then, by Theorem 42, $z^{*}\in\lim\mathcal{Z}(f_{m})$. This implies that

$\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}\subseteq\lim\mathcal{Z}(f_{m}).$

Let $z^{*}\in\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}^{c}$ be arbitrary.
Then $z^{*}=x+yi$ for some $x,y\in\mathbb{R}$ with $x\neq 0$. There are 2
cases: $x<0$ or $x>0$. If $x<0$, notice that

$|x-1|=-x+1>-x=|x|,$

and that

$|x-1|=-x+1>x+1,$

and that

$|x-1|=-x+1>-x-1.$

Consequently

$|x-1|>|x+1|.$

This implies that

$\displaystyle\beta_{1}(z^{*})=$ $\displaystyle|x-1+yi|$ $\displaystyle=$
$\displaystyle\sqrt{|x-1|^{2}+y^{2}}$ $\displaystyle>$
$\displaystyle\sqrt{|x|^{2}+y^{2}}$ $\displaystyle=$ $\displaystyle|x+yi|$
$\displaystyle=$ $\displaystyle|\beta_{2}(z^{*})|$

and that

$\displaystyle\beta_{1}(z^{*})=$ $\displaystyle|x-1+yi|$ $\displaystyle=$
$\displaystyle\sqrt{|x-1|^{2}+y^{2}}$ $\displaystyle>$
$\displaystyle\sqrt{|x+1|^{2}+y^{2}}$ $\displaystyle=$ $\displaystyle|x+1+yi|$
$\displaystyle=$ $\displaystyle|\beta_{3}(z^{*})|.$

Since $\alpha_{1}(z^{*})=1\neq 0$, Theorem 42 induces that
$z^{*}\in(\lim\mathcal{Z}(f_{m}))^{c}$.

If $x>0$, notice that

$|x+1|=x+1>x=|x|.$

Similarly, observe that

$|x+1|=x+1>x-1,$

and that

$|x+1|=x+1>-x+1,$

and consequently

$|x+1|>|x-1|.$

This implies that

$\displaystyle\beta_{3}(z^{*})=$ $\displaystyle|x+1+yi|$ $\displaystyle=$
$\displaystyle\sqrt{|x+1|^{2}+y^{2}}$ $\displaystyle>$
$\displaystyle\sqrt{|x|^{2}+y^{2}}$ $\displaystyle=$ $\displaystyle|x+yi|$
$\displaystyle=$ $\displaystyle|\beta_{2}(z^{*})|$

and that

$\displaystyle\beta_{3}(z^{*})=$ $\displaystyle|x+1+yi|$ $\displaystyle=$
$\displaystyle\sqrt{|x+1|^{2}+y^{2}}$ $\displaystyle>$
$\displaystyle\sqrt{|x-1|^{2}+y^{2}}$ $\displaystyle=$ $\displaystyle|x-1+yi|$
$\displaystyle=$ $\displaystyle|\beta_{1}(z^{*})|.$

Since $\alpha_{3}(z^{*})=1\neq 0$, by Theorem 42, induces that
$z^{*}\in(\lim\mathcal{Z}(f_{m}))^{c}$.

Thus for both cases $x>0$ and $x<0$, we obtain
$z^{*}\in(\lim\mathcal{Z}(f_{m}))^{c}$. This implies that
$\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}^{c}\subseteq(\lim\mathcal{Z}(f_{m}))^{c}$.
It follows that

$\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}\supseteq\lim\mathcal{Z}(f_{m}).$

Recall that
$\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}\subseteq\lim\mathcal{Z}(f_{m})$.
So we may conclude that

$\\{z\in\mathbb{C}:\operatorname{Re}(z)=0\\}=\lim\mathcal{Z}(f_{m}).$

When we work on our main theorem, we will show that the set of points from the
circle is $\lim\mathcal{Z}(P_{m})$. We will show this except for one limit
point of the circle. To show that the limit point is included in
$\lim\mathcal{Z}(P_{m})$, we need the following proposition.

######  Proposition 44.

Suppose $\\{f_{m}\\}_{m\geq 0}$ analytic in a region $D$. Then
$\underline{\lim}\,\mathcal{Z}(f_{m})$ is closed.

###### Proof.

Let $x^{\prime}$ be an arbitrary limit point of
$\underline{\lim}\,\mathcal{Z}(f_{m})$. Let $r>0$ be arbitrary. Given
$x^{\prime}$ is a limit point of $\underline{\lim}\,\mathcal{Z}(f_{m})$, it
follows that

$B\left(\frac{r}{2},x^{\prime}\right)\cap\underline{\lim}\,\mathcal{Z}(f_{m})\neq\emptyset.$

Then there exists $x^{*}\in
B\left(\frac{r}{2},x^{\prime}\right)\cap\underline{\lim}\,\mathcal{Z}(f_{m})$.
Given $x^{*}\in\underline{\lim}\,\mathcal{Z}(f_{m})$, there exists
$N\in\mathbb{N}$ such that, for any $n>N$,
$B\left(\frac{r}{2},x^{*}\right)\cap\mathcal{Z}(f_{n})\neq\emptyset$. Hence,
for arbitrary $m>N$, there exists $x_{m}\in
B\left(\frac{r}{2},x^{*}\right)\cap\mathcal{Z}(f_{m})$. It induces that
$|x_{m}-x^{*}|<\frac{r}{2}$. Recall that $x^{*}\in
B\left(\frac{r}{2},x^{\prime}\right)\cap\underline{\lim}\,\mathcal{Z}(f_{m})$,
which implies that $|x^{*}-x^{\prime}|<\frac{r}{2}$. Then we may deduce that

$\displaystyle|x^{\prime}-x_{m}|=$
$\displaystyle|x^{\prime}-x^{*}+x^{*}-x_{m}|$ $\displaystyle\leq$
$\displaystyle|x^{\prime}-x^{*}|+|x^{*}-x_{m}|$ $\displaystyle<$
$\displaystyle\frac{r}{2}+\frac{r}{2}$ $\displaystyle=$ $\displaystyle r$

So $x_{m}\in B(r,x^{\prime})$. Recall that $x_{m}\in\mathcal{Z}(f_{m})$. Then
$x_{m}\in B\left(r,x^{\prime}\right)\cap\mathcal{Z}(f_{m})$. Thus
$B\left(r,x^{\prime}\right)\cap\mathcal{Z}(f_{m})\neq\emptyset$. Since we
chose $m$ arbitrarily, it holds that

$\forall m>N,B\left(r,x^{\prime}\right)\cap\mathcal{Z}(f_{m})\neq\emptyset.$

Here, for arbitrary $r>0$, we found $N\in\mathbb{N}$ such that $\forall
m>N,B\left(r,x^{\prime}\right)\cap\mathcal{Z}(f_{m})\neq\emptyset$. That is,
$x^{\prime}\in\underline{\lim}\,\mathcal{Z}(f_{m})$. ∎

######  Lemma 45.

Let $a,b,c\in\mathbb{R}\backslash\\{0\\}$ where $b^{2}-4ac>0$. Suppose
$\\{P_{m}(z)\\}_{m\geq 0}$ is a sequence of functions of $z$ generated as
follows:

$\sum_{m=0}^{\infty}P_{m}(z)t^{m}=\frac{1}{(at^{2}+bt+c)(1-tz)}.$

Then there exists $N\in\mathbb{N}$ such that

$\mathcal{Z}(P_{m})\cap
B\left(\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right),0\right)=\emptyset,$

for all $m>N$, where $\alpha,\beta$ are the zeros of $at^{2}+bt+c$ with
$|\alpha|\leq|\beta|$.

###### Proof.

There are two cases: $ac>0$ or $ac<0$. Let us first prove the case $ac>0$.
Choose $z_{0}\in
B\left(\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right),0\right)$
be arbitrary. Then

$|z_{0}|<\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right).$

In view of $|\alpha|\leq|\beta|$, we may infer that

$\frac{1}{|\alpha|}+\frac{1}{|\beta|}\leq\frac{2}{|\alpha|}.$

It follows that

$\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right)\leq\frac{1}{|\alpha|}.$

Consequently, we obtain

$|z_{0}|<\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right)\leq\frac{1}{|\alpha|}.$

Theorem 36 implies that all the zeros of $P_{m}(z)$ are outside of the circle
$|z|=\frac{1}{|\alpha|}$, for all $m=1,2,\dots$. Hence $z_{0}$ is not a zero
of $P_{m}(z)$, for all $m=1,2,\dots$. This implies that

$z_{0}\notin\bigcup_{m=0}^{\infty}\mathcal{Z}(P_{m}).$

It concludes that

$\mathcal{Z}(P_{m})\cap
B\left(\frac{1}{2}\left(\frac{1}{|\alpha|}+\frac{1}{|\beta|}\right),0\right)=\emptyset,$

for all $m\geq 0$.

Now let us prove the case $ac<0$. We will consider $\\{H_{m}(z)\\}_{m\geq 0}$,
the sequence of functions of $z$ generated as follows:

$\sum_{m=0}^{\infty}H_{m}(z)t^{m}=\frac{1}{\left(\frac{ac}{|ac|}t^{2}+\frac{b}{\sqrt{|ac|}}t+1\right)(1-tz)}=\frac{1}{\left(-t^{2}+\frac{b}{\sqrt{-ac}}t+1\right)(1-tz)}.$

Let us denote those zeros of $-t^{2}+\frac{b}{\sqrt{-ac}}t+1$ by $t_{1},t_{2}$
where $|t_{1}|\leq|t_{2}|$.

We are to prove that $|t_{1}|\neq|t_{2}|$. Suppose, by contradiction, that
$|t_{1}|=|t_{2}|$. Since $t_{1},t_{2}$ are the zeros of
$\frac{ac}{|ac|}t^{2}+\frac{b}{\sqrt{|ac|}}t+1$, by Vieta’s Formula,
$t_{1}t_{2}=-1$. Then $|t_{1}|=|t_{2}|$ implies that

$1=|t_{1}t_{2}|=|t_{1}||t_{2}|=|t_{1}|^{2}.$

This induces that $|t_{1}|=|t_{2}|=1$. For the case $ac<0$, we may infer that

$\left\\{\begin{array}[]{cc}t_{1}=&1\\\ t_{2}=&-1\end{array}\right.\text{ or
}\left\\{\begin{array}[]{cc}t_{1}=&-1\\\ t_{2}=&1\end{array}\right.$

Notice that, by Vieta’s Formula,

$-\frac{b}{\sqrt{ac}}=t_{1}+t_{2}=0.$

This implies that $b=0$. It contradicts against our condition $b\neq 0$. Thus
our contrary supposition is false. So it holds that $|t_{1}|\neq|t_{2}|$.
Given that $|t_{1}|\leq|t_{2}|$, it follows that

$|t_{1}|<|t_{2}|.$

With Proposition 27 and Lemma 30, we may infer that, for
$z\neq\left\\{0,\frac{1}{t_{1}},\frac{1}{t_{2}}\right\\}$,

$H_{m}(z)=\frac{t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})t_{1}^{m+1}t_{2}^{m+1}z^{m+2}}{-(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)t_{1}^{m+1}t_{2}^{m+1}},m\geq
0.$

Vieta’s Formula induces $t_{1}t_{2}=-1$, which yields that

$H_{m}(z)=\frac{t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})(-1)^{m+1}z^{m+2}}{-(t_{2}-t_{1})(t_{1}z-1)(t_{2}z-1)(-1)^{m+1}},m\geq
0.$

Our plan is to apply Rouché’s Theorem to prove that there exists
$N\in\mathbb{N}$ such that the numerator of $H_{m}(z)$ has only one zero
inside the circle $|z|=\frac{|t_{1}|+|t_{2}|}{2}$ for all $m>N$. Define

$\displaystyle p(z)=$ $\displaystyle
t_{2}^{m+1}-t_{1}^{m+1}+(t_{1}^{m+2}-t_{2}^{m+2})z+(t_{2}-t_{1})(-1)^{m+1}z^{m+2}$
$\displaystyle q(z)=$ $\displaystyle-
t_{2}^{m+1}+t_{1}^{m+1}-(t_{1}^{m+2}-t_{2}^{m+2})z.$

Notice that the zero of $q(z)$ is

$\frac{-t_{2}^{m+1}+t_{1}^{m+1}}{t_{1}^{m+2}-t_{2}^{m+2}}.$

The triangle inequality implies that

(0.53)
$\left|\frac{-t_{2}^{m+1}+t_{1}^{m+1}}{t_{1}^{m+2}-t_{2}^{m+2}}\right|\leq\frac{|t_{2}|^{m+1}+|t_{1}|^{m+1}}{|t_{2}|^{m+2}-|t_{1}|^{m+2}}.$

Since $|t_{1}t_{2}|=|-1|=1$ and $|t_{1}|<|t_{2}|$, we may infer that
$|t_{1}|<1$ and $|t_{2}|>1$. From $|t_{2}|>1$, it follows that

(0.54) $|t_{2}|^{2}-1>0.$

From $|t_{1}|<1$, it follows the existence of $N_{1}\in\mathbb{N}$ such that

$3|t_{1}|^{m+1}+|t_{1}|^{m+3}<|t_{2}|^{2}-1,$