Let $G\subseteq \mathbb{C}$ be a bounded domain with $0 \in G$ and $f:\bar{G}\rightarrow\mathbb{C}$ is continous. Furthermore let $f$ be holomorphic in $G$ and let $|f(z)| \geq e^{Re (z)}$ for all $z\in\partial G$ and $|f(0)|<1$. I've got to show that $f$ has a zero.
My attempt:
Suppose $f(z)$ has no root. So that $|f(z)|>0$ is always fullfilled. Since the conditions for the maximum modulus theorem are fullfilled we apply it to $f(z)$ and $\frac{1}{f(z)}$ (which is also holomorphic on $G$ as $f(z) \neq 0$)
That is:
$0 \leq |e^{Re (z)}|\leq |f(z)| \leq f(C_1)$ for some $C_1 \in \partial G$, and
$0 \leq | \frac{1}{f(z)}|\leq |\frac{1}{e^{Re (z)}}|\leq \frac{1}{f(C_2)}$ for some $C_2 \in \partial G$
Now in the second row we get for $z=0$:
$\frac{1}{f(0)}\leq 1$ which is contradiction to the assumption that $|f(0)|<1$.
Is this proof correct? I mean can $0 \in \partial G$ be true when only $0\in G$ is assumed? Thank you for your help.
I don't think that your proof is correct. The inequality $|f(z)| \geq e^{\operatorname{Re} (z)}$ holds only on the boundary of $G$, but not inside $G$. (If it did then $f$ could not have a zero in the domain.)
If one “sees” that $ e^{\operatorname{Re} (z)} = |e^z|$, so that the condition $|f(z)| \geq e^{\operatorname{Re} (z)}$ is equivalent to $|f(z)| \ge |e^z|$ or $\left|\frac{e^z}{f(z)} \right| \le 1$ then the proof becomes apparent:
Assume that $f$ has no zeros in $G$. Then $h(z) = \frac{e^z}{f(z)}$ is holomorphic in $G$. On the boundary we have $$ |h(z)| = \frac{e^{\operatorname{Re} (z)}}{|f(z)|} \le 1 \, , $$ so that $|h(z)| \le 1$ for all $z \in G$. On the other hand, $|h(0)| = \frac{1}{|f(0)|} >1$. This is a contradiction, therefore $f$ must have a zero in $G$.