# Maximum modulus principle by Open Mapping theorem

This might be a duplicate question or straightforward one but I just need to clear it out. If I consider the version of the Maximum modulus theorem: "Let $f$ be a function analytic in a region $D$ and $|f(a)| > |f(z)|$ for all $z \in D,$ then $f$ is a constant."

My approach: Suppose to the contrarty that $f$ is not constant. Then by open mapping theorem any neighborhood of $a,$ say $B_{\delta}(a)$ maps to an open neighborhood of $f(a),$ say $B_{\epsilon}(f(a)).$ How can I give a mathematically precise proof for this ? Thank you for your help in advance.

• The open mapping and the maximum modulus both follow from $f(z) = f(z_0) + C \ (z-z_0)^n + o(|z-z_0|^n)$ for some $C \ne 0$ whenever $f(z)$ is not constant – reuns Oct 5 '16 at 2:59
• $|f(a)|> |f(z)|$ for all $z\in D$ is impossible. – zhw. Oct 5 '16 at 3:09

Every open neighbourhood of a point contains points whose modulus is larger.

• Thanks. Could you please explain that a little bit. – user358174 Oct 4 '16 at 22:34
• @ManMath Draw a picture, and it should become clear. – mickep Oct 5 '16 at 14:46
• @mickep I've already done that, which is clear. I just want it to be mathematically precise. Anyways thanks everyone for the suggestions. As I said in my last comment, I'm trying to comprehend what Conway mentioned. – user358174 Oct 5 '16 at 15:13

The open mapping and the maximum modulus both follow from $f(z) = f(z_0) + C \ (z-z_0)^n + o(|z-z_0|^n)$ for some $C \ne 0$ and $n \in \mathbb{N}^*$ whenever $f(z)$ is holomorphic and non-constant.

• If $f(z)$ is holomorphic on $U$ then $\sup_{z \in U} |f(z)| = |f(a)|$ where $a \in U$ leads to $|f(a)| \ge |f(a+h)| = |f(a) + C \ h^n + o(|h|^n)|$. If $f(a) = 0$ this is clearly impossible if $C \ne 0$. So assume $f(a) \ne 0$, you get that $1 \ge \frac{|f(a+h)|}{|f(a)|} = |1 + \frac{C}{f(a)} \ h^n + o(|h|^n)|$ a contradiction if $C \ne 0$ by taking $h = \epsilon (\overline{C/f(a)})^{1/n}$ and letting $\epsilon \to 0^+$. Hence $C= 0$ and it means $f(z)$ is constant.

• Assume that $f(U)$ is not open, then there is a point $b \in \partial f(U)$ such that $a \in U, b = f(a)$. A contradiction again if $f$ is non-constant since $f(a+h) = f(a)+ C h^n + o(|h|^n)$ so $f(a+B_\epsilon(0))$ contains an open ball (*) around $f(a)$ whenever $C \ne 0$.

(*) contradicting that $f(a) \in \partial f(U)$, $a \in U$

• user1952009 thank you. I'm trying to understand Conway's proof which is much shorter. Still can't comprehend it. – user358174 Oct 5 '16 at 13:43
• @ManMath I don't think it is shorter, and fundamentally it can't be different to thisone – reuns Oct 5 '16 at 13:49
• apparently it is. (pg. 128). It's ok I'll go through it again. – user358174 Oct 5 '16 at 13:53