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Let $f$ be an analytic function on $|z|\leq 1$ with $f(0)=0$ and let $|f(z)|$ have a maximum for $|z|\leq 1$ at $z_0=1$. Show that $f^{'}(z_0)\neq 0$ unless $f$ is constant.

Here is what I have: If $|f(1)|=0$ then $f$ is constant and we are done. So assume $|f(1)|>0$. Then $f(1)\neq 0$ and the function $g(z)=\frac{f(z)}{f(1)}$ is analytic. Also $g$ satisfies the conditions of Schwarz Lemma.

Here is where I am stuck. I want to get $g(z)=cz$ for some $c$ with $|c|=1$.

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Theorem 4.1 in this link may be useful –  PAD Jul 29 '12 at 21:55

1 Answer 1

By "$f$ is an analytic function on $|z|\le 1$" I assume you mean $f$ is analytic on some neighbourhood of $|z|\le 1$. If $f$ is not constant, there is some smallest positive integer $k$ such that $f^{(k)}(1) \ne 0$. Suppose $f'(1) = 0$, so $k \ne 1$. Thus $f(z) = f(1) + a (z-1)^k + O(|z-1|^{k+1})$ as $z \to 1$, with $a \ne 0$, $f(1) \ne 0$ and $k \ge 2$. So $|f(z)|^2 = |f(1)|^2 + 2 \text{Re}(\overline{f(1)} a (z-1)^k) + O(|z-1|^{k+1})$. As $\omega$ goes around the unit circle, $\overline{f(1)} a \omega^k$ makes $k$ revolutions around the origin. There is some $\omega$ in the open left half plane such that $\text{Re}(\overline{f(1)} a \omega^k) > 0$. Then for $z = 1 + s \omega$, if $s > 0$ is sufficiently small we have $|f(z)| > |f(1)|$ and $|z| < 1$, contradicting the maximality of $f$ at $z=1$.

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