Schwarz Lemma states the following:

Let $D = \{z : |z| < 1\}$ be the open unit disk in the complex plane centered at the origin and let $f : D \to D$ be a holomorphic map such that $f(0) = 0$.

Then, $|f(z)| \le |z|$ for all $z$ in D and $|f'(0)| ≤ 1.$

Now $f(z)=e^{z}-1$ is holomorphic because $f'(z)=e^{z}$ and $f(0)=0$. Then $f$ must satisfy the Lemma, but I can't see how this is the case as


Of course there must be something I am not understanding.

  • 3
    $\begingroup$ $f(D)\not\subseteq D$. $\endgroup$ – David Mitra Feb 20 '15 at 11:48

$f$ must also satisfy: $\forall_{z\in D}|f(z)| \le 1$.

  • 1
    $\begingroup$ That's (almost) what it means when it says $f : D \to D$. $\endgroup$ – GEdgar Feb 20 '15 at 14:13

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