If $f$ is holomorphic and satisfies $f(0)=0$ and $|f(z)| \leq \exp[(\Re z)^2 - (\Im z)^2]$, then $f=0$ If $f$ is holomorphic and satisfies $f(0)=0$ and $|f(z)| \leq \exp[(\Re z)^2 - (\Im z)^2]$, then $f=0$

I thought about using uniqueness, though I have a problem showing that I have a limit point of $0$.
I can show that I have a limit point of epsilon for any epsilon that I want, but I cannot show that there is even a single $z f(z)=0$ beside $z=0$.
Just to complete the story in the comments, one has $$\left|\frac{f(z)}{e^{z^2}}\right|\leq e^{\mathfrak{(R\mathrm z)^{2}-(\mathfrak{I}\mathrm{z})^{2}}}\cdot \frac 1{\left|e^{z^2}\right|}=e^{\mathfrak{R}(z^2)}\cdot\frac 1{e^{\mathfrak{R}(z^2)}}=1,$$ which implies $$\frac{f(z)}{e^{z^2}}=C,$$ by Liouville. Now $f(0)=0$ shows that $C=0.$ It follows that $f(z)=0.$ QED
Actually I understood it quite differently :
$\begin{array}{c}
|f(z)|\leq e^{(\Re z)^{2}-(\Im z)^{2}}=e^{\Re z^{2}}=|e^{z^{2}}|\\
0\leq\frac{|f(z)|}{|e^{z^{2}}|}\leq1
\end{array}$
now by Liouville:
$
\begin{array}{c}
\frac{|f(z)|}{|e^{z^{2}}|}=c\\
f(z)=c|e^{z^{2}}|\\
0=f(0)=c\Rightarrow f(z)=0
\end{array}$