$\newcommand{\abs}[1]{\left|#1\right|}$
Solution-HW2.pdf
### Chapter 3.1

Find the radius of convergence of each of the following power series:

(a) $\sum_{n=0}^∞a^n z^n, a \in \mathbb{C}$

Solution: To find the radius of convergence $R$ of a power series $\sum_{n=0}^∞a_n z^n$, we generally need to check $\frac{1}{R}=\limsup _{n→∞}\left|a_n\right|^{\frac1n}$. Of course, here $\left|a_n\right|^{\frac1n}$ is just the constant sequence $\abs{a}$, so $R=\frac{1}{\abs{a}}$ (ie, $∞$ if $\left.a=0\right)$. We could also have found this by noting that this series is just a change of variables $z \mapsto a z$ of a geometric series, so it converges when $\abs{a z}< 1$.
(b) $\sum_{n=0}^∞a^{n^2} z^n, a \in \mathbb{C}$

Solution: In this case, $\left|a_n\right|^{\frac1n}=\abs{a}^n$. Therefore, the radius of convergence depends on $a$, as follows:
- if $\abs{a}< 1$, then $\limsup _{n→∞}\abs{a}^n=0$, so $R=∞$.
- if $\abs{a}=1$, then we are just taking the limsup of $\abs{a}^n=1$, so $R=1$.
- if $\abs{a}>1$, then $\limsup _{n→∞}\abs{a}^n=∞$, and $R=0$.

(d)\[
\sum_{n=0}^∞z^{n !}
\]
Solution: Here for $n>1,\left|a_n\right|^{\frac1n}=1$ if $n=k!$ for some non-negative integer $k$, and 0 otherwise. This means that
\begin{aligned}
\limsup _{n→∞}\left|a_n\right|^{\frac1n} &=\lim _{n→∞} \sup _{m \geq n}\left|a_m\right|^{\frac{1}{m}} \\
&=\lim _{n→∞} \sup \{0,1\},
\end{aligned}
and so the radius of convergence is also 1.

### Chapter 3.2

Show that $f(z)=\abs{z}^2=x^2+y^2$ has derivative only at the origin.

Solution: Suppose that $f$ is differentiable at some $a \in \mathbb{C}$ ie, that the limit
\[
L=\lim _{h→0} \frac{f(a+h)-f(a)}{h}
\]
exists. Consider the line $\ell$ between the origin and $a$, and take its perpendicular $\ell^{\perp}$ through $a$. Let $a_t$ be the point on $\ell$ at distance $t$ further from the origin than $a$, and $a_t^{\perp}$ be either point on $\ell^{\perp}$ at distance $t$ from $a$. On the one hand, $\left|a_t\right|=\abs{a}+t$ and $h_t:=a_t-a=t e^{i θ}$, where $θ=\arg a$, so
\begin{aligned}
\lim _{h_t→0} \frac{f\left(a_t\right)-f(a)}{h_t} &=\lim _{t→0} \frac{\abs{a}^2+2\abs{a} t+t^2-\abs{a}^2}{t e^{i θ}} \\
&=\lim _{t→0} \frac{2\abs{a}+t}{e^{i θ}} \\
&=\frac{2\abs{a}}{e^{i θ}}
\end{aligned}
On the other, by the Pythagorean theorem $\left|a_t^{\perp}\right|^2=\abs{a}^2+t^2$, while $h_t^{\perp}:=a_t^{\perp}-a=t e^{i\left(θ±\fracπ{2}\right)}$, meaning
\begin{aligned}
\lim _{h_t^{\perp}→0} \frac{f\left(a_t^{\perp}\right)-f(a)}{h t^{\perp}}&=\lim _{t→0} \frac{\abs{a}^2+t^2-\abs{a}^2}{t e^{i\left(θ±\fracπ{2}\right)}}\\
&=\lim _{t→0}\frac{t}{e^{i\left(θ±\fracπ{2}\right)}}\\&=0
\end{aligned}
But in both cases were are taking a limit of the difference quotient as $z$ approaches $a$, so they must both be equal to $L$. This is a contradiction unless $a=0$, so it remains merely to check that $f$ actually has a derivative at 0. This is easy, since in that case
\begin{aligned}
\lim _{h→0} \frac{f(h)-f(0)}{h} &=\lim _{h→0} \frac{\abs h^2}{h} \\
&=\lim _{h→0} \bar{h} \\
&=0
\end{aligned}