Let $f$ be analytic in in $|z| \leq 1$ and let $|f(z)|$ be maximized for $|z| \leq 1$ at $z_0$ with $|z_0|=1$. Prove that $|f'(z_0)| \neq 0$ unless $f$ is a constant.
My approach: By the maximum modulus theorem, $|f(z_0)| \geq |f(z)|$ for all $|z| \leq 1.$ This yields $|f'(z_0)| \geq 0$ in $|z| \leq 1.$ I don't know how to complete the problem. Any help is appreciated.
Let $D=\{z: |z|\leq 1\}.$ WLOG , for easier calculation assume $f(z_0)\in \mathbb R^+.$ Suppose $f$ is not constant on $D$.
(1). By contradiction, suppose $f'(z_0)= 0.$ Then for some $n\in \mathbb N $ with $n\geq 2$ we have $f(z)=f(z_0)+(z-z_0)^ng(z)$ for all $z\in D,$ where $g$ is analytic on $D$ and $g(z_0)\ne 0.$ And for $z\in D$ we have $g(z)=g(z_0)(1+h(z))$ where $\lim_{z\to z_0; z\in D}|h(z)|=0.$
(2).Since $n\geq 2$ and $g(z_0)\ne 0,$ if $U$ is any open subset of $\mathbb C$ with $z_0\in U,$ there exists $z_U\in (U\cap D)$ \ $\{z_0\}$ such that $(z_U-z_0)^ng(z_0)\in \mathbb R^+.$
(3).Now let $U$ be an open disc centered at $z_0$, with small enough radius that $z\in U\cap D\implies |h(z)|<1/2.$ So $|h(z_U|<1/2.$ We have $$f(z_U)=f(z_0)+d+h(z_U)d$$ where $d=(z_U-z_0)^n(z_U-z_0)^n>0.$ Recalling that $f(z_0)>0$ we have $$|f(z_U)|\geq |f(z_0)+d|-|h(z_U)d|=f(z_0)+d-|h(z_U)d|> f(z_0) +d/2>f(z_0)=|f(z_0)|,$$ contrary to $|f(z_0|=\max \{|f(z)|:z\in D\}.$
Remark: Step (2) is valid because $n\geq 2.$ It would not be valid for $n=1.$