# Application of maximum modulus theorem to $f(z)$ and $\frac{1}{f(z)}$

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.

• What do you mean by "$f$ has a root"? $f$ is a function not an equation. – Empty Aug 14 at 16:30
• Of which inequality? Well, I've got to show that f has a zero (or root). – Thesinus Aug 14 at 16:40
• I forget the exact question but I will look for it if needed but someone said that for a 0 solution in these cases you can use SOLUTION(X) - SOLUTION(X) ... – Jay Aug 14 at 16:41

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 $$G$$.
• Thank you for your genial post! To my attempt: Doesn't a function always 'send' an argument to just 'one' value. I mean if $f(0)<1$ for $0 \in G$ then it always has to be true no matter where $0$ is. Of course by hypothesis, we can assume $0\in \partial G$or not?, – Thesinus Aug 14 at 16:56
• @Thesinus: I do not understand what you are saying, to be honest. But $0\in \partial G$ is not possible, because then $1 > |f(0)| \ge e^{\operatorname{Re} 0} = 1$. – Martin R Aug 14 at 16:59
• @Thesinus: It is also given that $0 \in G$, that excludes the possibility that $0$ is on the boundary of $G$: For an open set, $G$ and $\partial G$ are disjoint. – Martin R Aug 14 at 17:08