Note: This is a problem from an introduction course in complex analysis for engineers.
Problem: Show that if $f$ is analytic when $z \neq 0$ and $|f(z)| \leq \sqrt{|z|} + \frac{1}{\sqrt{|z|}} $ then $f$ is a constant.
Given: $f(z) = \sum_{n=-\infty}^{\infty} c_n z^n $ in the plane $ 0 < |z| < \infty$.
Attempt at solution: By the cauchy inequality theorem we have \begin{equation}C_n \leq \frac{\sqrt{p} + \frac{1}{\sqrt{p}}}{p^n} \end{equation}
where $p$ is the radius of the circle $|z| = p > 0$ in the complex plane and $n \in \mathbb{Z} $.
We want to show that $C_n = 0 $ for $n \geq 1 $ and $n \leq -1$, because this would mean $f = C_o = constant.$ We see that if $p \rightarrow \infty $ and $n \geq 1$ then \begin{equation}C_n \leq \lim_{p\to\infty} \frac{\sqrt{p} + \frac{1}{\sqrt{p}}}{p^n} = 0. \end{equation}
On the other hand, if $p \rightarrow \infty $ and $n \leq -1$ then
\begin{equation}C_n \leq \lim_{p\to\infty} \frac{\sqrt{p} + \frac{1}{\sqrt{p}}}{p^n} = \infty. \end{equation}
The first limit gives us $C_n = 0 $ for $n \geq 1$, great. The second limit doesn't give us what we want (namely $C_n = 0$.). I am stuck here trying to prove that $C_n = 0$ for $n \leq -1$. Any help appreciated.
