$f$ is a complex constant function

Question

If $f$ is analytic in a domain $D$ and is there exists $z_0 \in D$ such that $|f(z)|\geq |f(z_0)|>0$ is true for every $z\in D,$ then $f$ is constant.

It's kind of strange this result because it's like the Maximun Principle, but the inequality is backwards... Is possible this result?

I tried to prove that if $f=u+iv,$ then $u$ is bounded below, but I got $|f(z_0)|\leq |u|.$ Thanks for your help!

Answer 1

So $1/f$ is analytic in the domain and attaining the upper bound in the domain.

Answer 2

Let $g=1/f$. Then $g$ is analytic in $D$ since $f\not=0$ in $D$. Moreover, $|g(z)|=1/|f(z)|\leq 1/|f(z_{0})|=|g(z_{0})|$ for all $z\in D$. Thus, $g$ is constant by the Maximum Modulus Principle and, hence, $f$ is constant.