Show $f\left(z\right)=\sum_{n=0}^{\infty}\frac{z^{n}}{a^{n^{2}}}$ has exactly one zero in a certain domain. I'm studying for an exam and I can't manage to solve the following practice question: Let $a$
  be constant with $\left|a\right|>10$
 . We define an entire function $f$
  by: $$f\left(z\right)=\sum_{n=0}^{\infty}\frac{z^{n}}{a^{n^{2}}}$$
 Show that for all $m\in\mathbb{N}$
  the function $f$
  has exactly one zero in the ring $$\mathcal{A}_{m}:=\left\{ \left|a\right|^{2m-2}<\left|z\right|<\left|a\right|^{2m}\right\} $$
This obviously screams to use Rouche's Theorem somehow but I can't figure out how.
Help would be appreciated
 A: Yes, Rouche!
Let $$f_m(z)=\sum_{n=0}^m\frac{z^n}{a^{n^2}}.$$
Then for $|z|=a^{2m}$ we have
$$ \left|f_m(z)-f_{m-1}(z)\right|=\frac{(a^{2m})^m}{a^{m^2}}=a^{m^2}$$
and 
$$ \left|f_{m-1}(z)\right|\le\sum_{n=0}^{m-1}\frac{a^{2nm}}{a^{n^2}}=\sum_{n=0}^{m-1}a^{n\cdot(2m-n)}\le \sum_{n=0}^{m-1}a^{m^2-m+n}<a^{m^2}\cdot\sum_{k=1}^\infty a^{-k}=a ^{m^2-1}.$$
Hence $f_m$ and $\frac{z^m}{a^{m^2}}$ have the same numberof roots in $\{|z|<a^{2m}\}$.
Also for $|z|=a^{2m}$, 
$$ |f(z)-f_m(z)|\le \sum_{n=m+1}^\infty \frac{a^{2mn}}{a^{n^2}}=\sum_{n=m+1}^\infty a^{(2m-n)n}<\sum_{k=1}^\infty a^{m^2-k}=a^{m^2-1}$$
and 
$$ |f_m(z)|\ge |f_m(z)-f_{m-1}(z)|-|f_{m-1}(z)|\ge a^{m^2}-a^{m^2-1}>a^{m^2-1}$$
imply that $f$ also has $m$ roots in the open $2m$-disk. By the same inequalities, $f$ has no roots on the $2m$-circle.
A: Using the following makes the calculations above probably a lot shorter:
Considering $m=1$: saying that $f$ has a zero in $1 < | z | < | a |^{2}$ is equivalent to saying that $f$ has a zero in $| a |^{2m-2} < | a^{2m-2} z | < | a |^{2m}$. Using this we see that saying that $f(z)$ has a zero in $\mathcal{A}_{m}$ is equivalent to saying that $f(a^{2-2m} z)$ has precisely one zero in $\mathcal{A}_{1}$ for all $m$.
