$f(z)$, a function that is analytic in field $D$

Asked Mar 04 '23 at 19:17

Active Mar 04 '23 at 19:26

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Let there be $f(z)$, a function that is analytic in field $D$. Prove that if $|f(z)|=\text{const}$ for every $z$ in $D$, then the function $f(z)$ is constant in $D$. field $D$ is connected and opened disk *

There is a solution by our university's prof, but I think it's wrong, solution: We define a new function $g(z) = \frac{f(z)}{e^z - 1}$

We note that $g(z)$ is analytic for every $z$ except $z=2\pi ni$ for some integer $n$. We also pay attention that $g(z) \leq 1$ for every neighborhood around those singular points. to be honet I didn't get it why $g(z) \leq 1$ , and if may someone help me in a direction I would be thankful

edited Mar 04 '23 at 19:26

asked Mar 04 '23 at 19:17

mayyas haj

What do you mean by field $D$? – Koro Mar 04 '23 at 19:22
We call a field as a simple opened disk – mayyas haj Mar 04 '23 at 19:23
So $D$ is open and connected. Isn't it? – Koro Mar 04 '23 at 19:24
I am sorry I said it's simple , it's opened and connected you are correct! – mayyas haj Mar 04 '23 at 19:26
Will you accept any other solution (one that doesn't involve defining $g$)? – Koro Mar 04 '23 at 19:27
of course brother anything – mayyas haj Mar 04 '23 at 19:33
Then here is a hint for you: Suppose that $f=u+iv$ and use Cauchy Riemann's equations. – Koro Mar 04 '23 at 19:36
thank you so much – mayyas haj Mar 04 '23 at 19:36
Another approach: If we assume that $f$ is non constant, then the image of $D$ under $f$ must be an open set by open mapping thm. But $|f|\equiv \text{constant}$ means the image of $D$ is a connected subset of some circle centered around the origin, which is a contradiction. So, $f$ must be constant. – Matthew H. Mar 04 '23 at 19:40

$f(z)$, a function that is analytic in field $D$

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