A particle is located on the coordinate plane at $(5,0)$. Define a ''move'' for the particle as a counterclockwise rotation of $\frac{\pi}{4}$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction.  Find the particle's position after $150$ moves.
Let $z_0 = 5,$ and let $z_n$ be the position of the point after $n$ steps.  Then
\[z_n = \omega z_{n - 1} + 10,\]where $\omega = \operatorname{cis} \frac{\pi}{4}.$  Then
\begin{align*}
z_1 &= 5 \omega + 10, \\
z_2 &= \omega (5 \omega + 10) = 5 \omega^2 + 10 \omega + 10, \\
z_3 &= \omega (5 \omega^2 + 10 \omega + 10) + 10 = 5 \omega^3 + 10 \omega^2 + 10 \omega + 10,
\end{align*}and so on.  In general, we can prove by induction that
\[z_n = 5 \omega^n + 10 (\omega^{n - 1} + \omega^{n - 2} + \dots + 1).\]In particular,
\[z_{150} = 5 \omega^{150} + 10 (\omega^{149} + \omega^{148} + \dots + 1).\]Note that $\omega^4 = \operatorname{cis} \pi = -1$ and $\omega^8 = 1.$  Then by the formula for a geometric series,
\begin{align*}
z_{150} &= 5 \omega^{150} + 10 (\omega^{149} + \omega^{148} + \dots + 1) \\
&= 5 \omega^{150} + 10 \cdot \frac{1 - \omega^{150}}{1 - \omega} \\
&= 5 (\omega^8)^{18} \cdot \omega^6 + 10 \cdot \frac{1 - (\omega^8)^{18} \cdot \omega^6}{1 - \omega} \\
&= 5 \omega^6 + 10 \cdot \frac{1 - \omega^6}{1 - \omega} \\
&= 5 \omega^6 + 10 (\omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) \\
&= -5 \omega^2 + 10 (-\omega - 1 + \omega^3 + \omega^2 + \omega + 1) \\
&= 10 \omega^3 + 5 \omega^2 \\
&= 10 \operatorname{cis} \frac{3 \pi}{4} + 5i \\
&= 10 \cos \frac{3 \pi}{4} + 10i \sin \frac{3 \pi}{4} + 5i \\
&= -5 \sqrt{2} + (5 + 5 \sqrt{2}) i.
\end{align*}Thus, the final point is $\boxed{(-5 \sqrt{2}, 5 + 5 \sqrt{2})}.$
Let $z_0 = 5,$ and let $z_n$ be the position of the point after $n$ steps.  Then
\[z_n = \omega z_{n - 1} + 10,\]where $\omega = \operatorname{cis} \frac{\pi}{4}.$  Then
\begin{align*}
z_1 &= 5 \omega + 10, \\
z_2 &= \omega (5 \omega + 10) = 5 \omega^2 + 10 \omega + 10, \\
z_3 &= \omega (5 \omega^2 + 10 \omega + 10) + 10 = 5 \omega^3 + 10 \omega^2 + 10 \omega + 10,
\end{align*}and so on.  In general, we can prove by induction that
\[z_n = 5 \omega^n + 10 (\omega^{n - 1} + \omega^{n - 2} + \dots + 1).\]In particular,
\[z_{150} = 5 \omega^{150} + 10 (\omega^{149} + \omega^{148} + \dots + 1).\]Note that $\omega^4 = \operatorname{cis} \pi = -1$ and $\omega^8 = 1.$  Then by the formula for a geometric series,
\begin{align*}
z_{150} &= 5 \omega^{150} + 10 (\omega^{149} + \omega^{148} + \dots + 1) \\
&= 5 \omega^{150} + 10 \cdot \frac{1 - \omega^{150}}{1 - \omega} \\
&= 5 (\omega^8)^{18} \cdot \omega^6 + 10 \cdot \frac{1 - (\omega^8)^{18} \cdot \omega^6}{1 - \omega} \\
&= 5 \omega^6 + 10 \cdot \frac{1 - \omega^6}{1 - \omega} \\
&= 5 \omega^6 + 10 (\omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) \\
&= -5 \omega^2 + 10 (-\omega - 1 + \omega^3 + \omega^2 + \omega + 1) \\
&= 10 \omega^3 + 5 \omega^2 \\
&= 10 \operatorname{cis} \frac{3 \pi}{4} + 5i \\
&= 10 \cos \frac{3 \pi}{4} + 10i \sin \frac{3 \pi}{4} + 5i \\
&= -5 \sqrt{2} + (5 + 5 \sqrt{2}) i.
\end{align*}Thus, the final point is $\boxed{(-5 \sqrt{2}, 5 + 5 \sqrt{2})}.$
The final answer is $\[-5 \sqrt{2} + (5 + 5 \sqrt{2}) i$. I hope it is correct.