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Let $f$ be a one-to-one holomorphic mapping from the unit disk onto itself, $f(0)=0$, $f^{\prime}(0)>0$. Prove that $f(z)=z$.

Attempt:

Since the hypothesis gives us $f(0)=0$ and $|f(z)<1|$ (since $f(z)$ is mapped to the unit disk), we can apply Schwarz's lemma to $f$ and $f^{-1}$ to show that $|f(z)|\le |z|$ and that $|f(z)| \ge |z|$ respectively. This implies that $f(z)=e^{i\theta}z$ (again by Schwarz's lemma).

Now, how to get rid of the $e^{i\theta}$ in $f(z)=e^{i\theta}z$ using $f^{\prime}>0$? It seems that the Maximum Modulus Theorem may work here but I can't complete the idea.

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    $\begingroup$ @Bob first idea is right, just use $f'(0)>0$ $\endgroup$
    – Albert
    Apr 14, 2015 at 9:14

1 Answer 1

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$f'(z) = e^{i\theta}$, so if $f'(z) > 0$, we must have $\theta = 0$ (or $2\pi i k$ for some $k\in\mathbb{Z}$ if we want to pedantic).

(Note that $e^{i\theta}$ is some point on the unit circle. The only such point which is positive is $1$.)

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