# Prove $z\rightarrow \mid f(z)\mid$ cannot have a local min in $U\subset\mathbb{C}$ open subset

We discussed in class an example for the Maximum Modulus Principle:

Prove that if the subset $U\subset\mathbb{C}$ is open, connected, $f:U\rightarrow \mathbb{C}$ is holomorphic, non constant and has nonzeros in $U$, then $z\rightarrow \mid f(z)\mid$ cannot have a local min in $U$.

Taking $1/f$, it is well-defined since, by assumption, there are no zero values of $f$, but I missed the argument why this function cannot have a minimum. Could someone please explain?

• Have you looked at Wikipedia? en.wikipedia.org/wiki/Maximum_modulus_principle Nov 21, 2017 at 16:59
• You don't want to show that $1/f$ cannot have a minimum! You want to show $f$ cannot have a mimimum, which is the same as showing what about $1/f$? Nov 21, 2017 at 17:00

Note that the map $x\mapsto 1/x$ reverses the order of the interval $(0,\infty)$. That is, the statements $x<y$ and $\frac{1}x>\frac{1}y$ are equivalent. Thus, if you want to show that $f$ does not have a local minimum, you want to show that $1/f$ does not have a local maximum - but it sounds like you already have the maximum modulus principle, which exactly says that $1/f$ does not have a local maximum.
The open mapping theorem implies that $f(U)$ is open. If $f$ has no zeros in $U$ then $|f(U)|$ is an open interval that is a subset of $(0,\infty)$ and hence has no $\min$.
Note the distinction between $\min$ and $\inf$: For example, with $U=B(0,1)$ and $f(z) = 1-z$ that $|f(U)| = (0,1)$, that is, $\inf |f(U)| = 0$.