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Let $f$ be a non-constant analytic function in $G\subseteq \mathbb{C}$ (connected, bounded and open) and cont. in $\overline{G}$ such that

$$\exists \lambda >0, \text{ with } |f(z)| = \lambda, \forall z \in \partial G$$

Then $$\exists z \in G, \text{ with } f(z) = 0$$

I have two different approaches designed by me and I want to see if I got both right.

First:

Since $f$ is analytic in $G$ and continuous in $\overline{G}$ by the Maximum Modulus Theorem (Second version of Conway's book)

$$\max_{z \in \overline{G}}\{|f(z)|\} = \max_{z \in \partial G}\{|f(z)|\}$$

Then we have that $|f(z)| \leq \lambda, \forall z \in \overline{G}$.

Now consider the function

$$g(z) = \frac{1}{f(z)} $$

Assume $f(z) = 0$ has no solutions in $G$, then $g(z)$ is analytic in $G$ and continuous in $\overline{G}$ because $f$ is, also $g$ satisfies

$$|g(z)| = \frac{1}{|f(z)|} \leq \frac{1}{\lambda}, \forall z \in \overline{G}$$

By the Maximum modulus theorem (first version of Conway's book), there exists $z_m \in \overline{G}$ such that $|g(z_m)| = \frac{1}{\lambda} \geq g(z)$ for all $z \in \overline{G}$. Thus, $g$ is constant which implies that $f$ is constant and we get a contradiction.

Second proof:

Since $f$ is analytic in $G$ and continuous in $\overline{G}$ by the Maximum Modulus Theorem (Second version of Conway's book)

$$\max_{z \in \overline{G}}\{|f(z)|\} = \max_{z \in \partial G}\{|f(z)|\}$$

Also, $f$ is non-constant and by the Open Mapping Theorem, $f$ must map open sets to open sets, but the modulus of $f$ is bounded for every point in $G$, then $f(G) \subset \{z \in \mathbb{C} : |z| < \lambda\}$ and because $f$ is continuous in $\overline{G}$, then $f(\partial G)$ is the border of the disk or contained in it.Hence, the $\operatorname{Range}(f) = \{z \in \mathbb{C}: |z| \leq 1 \}$, and disk pass through the origin, and must exist a point in $G$ such that f(z) = 0.

I will want to make a better description of the second one, but I think that it is enough. Which one is better for a qualifying exam and if I have to fix some details please let me know.

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    $\begingroup$ By the MMP $|f(z)| \le \sup_{|s| \in \partial G} |f(s)| = 1$. If $f$ has no zero then $g(z) = \frac{1}{f(z)}$ is analytic and by the MMP $\frac{1}{|f(z)|} = |g(z)| \le \sup_{|s| \in \partial G} |g(s)| = 1$. Thus $|f| =1$ is constant and so is $f$ (why ?) $\endgroup$ – reuns Aug 26 '17 at 1:47
  • $\begingroup$ Is not by the maximum modulus theorem that says that if an analytic function is bounded by a value in the domain is constant? @reuns $\endgroup$ – Richard Clare Aug 26 '17 at 1:52
  • $\begingroup$ That's exactly what I prove in the first proof with $\lambda$ instead of $1$. @reuns $\endgroup$ – Richard Clare Aug 26 '17 at 1:53
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    $\begingroup$ Sure but it seems you think there are many different versions, while the core argument is that $f$ is analytic. Locally if $f$ is non-constant then $f(z+a) \sim f(a) + C z^n$ for some $C \ne 0, n \ge 1$. This proves the open mapping and the maximum modulus principle (there are no local maximum of $|f|$) $\endgroup$ – reuns Aug 26 '17 at 1:56
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First version is not accurate, specifically

... there exists $z_m\in \overline{D}$ such that $|g(z_m)|=\frac{1}{λ}\geq g(z)$ for all $z∈\overline{D}$. Thus, $g$ is constant ...

$\overline{D}$ is closed, MMP restricts the result to open sets, consequence of this is that $g$ reaches the maximum at the boundary $\partial G \subset \overline{D}$. So $z_m \in \partial G \subset \overline{D}$, no contradictions so far.

To fix the problem, you need to fix a point $z_0 \in D$ (open set!) for which $|f(z_0)|<\lambda$ (you already deducted this), thus $|g(z_0)|=\left|\frac{1}{f(z_0)}\right| > \frac{1}{\lambda} \geq |g(z)|$. So $\exists z_0 \in D$ such that $|g(z_0)|\geq |g(z)|, \forall z \in D$ and only now you can conclude that $g$ is constant in $D$.

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