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(1)Let $f$ be analytic on a convex open subset $\Omega$ of the complex plane $\mathbb C$ with $|f'| \le 1$ on $\Omega$. Prove that $|f(a)-f(b)|\le|a-b|$ whenever $a, b \in \Omega$.

(2)Let $a, b \in \{z:\mathrm {Re}(z)<0\}$. Show that $|e^a-e^b|<|a-b|$.

If $f$ is a function of real variable, (1) and (2) are straightforward by mean-value theorem. However, I can't apply mean-value theorem in this case because $f$ is of complex variable.

Searching at Wikipedia, I found the complex mean value theorem which state that there is a $t, s$ on a segment $[u,v]$ such that $\mathrm {Re}(f'(t))= \mathrm {Re}[ \frac {f(u)-f(v)}{u-v}]$ and $\mathrm {Im}(f'(s))=\mathrm {Im}[\frac {f(u)-f(v)}{u-v}]$, but this also doesn't work because it cannot guarantee that such $t$ and $s$ coincide, and also I can only get the information such that $|\mathrm {Re}[ \frac {f(u)-f(v)}{u-v}]| \le 1$ and the same for the imaginary part.

What should I do?

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    If $\gamma$ is the line segment from $a$ to $b$, then $|f(b)-f(a)| = |\int_{\gamma} f'(z)dz| \leq \int_{\gamma} |f'(z)| \; |dz|$. Use that. – 2017-02-14
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    $f(b)-f(a) = \int_a^b f'(z)dz$ (note this integral is well-defined whatever the path $a \to b$ because $f'$ is analytic) – 2017-02-14
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    Does this work as the same for (2)? – 2017-02-14
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    Oh, I found that it works as the same for both (1) and (2). Thanks. – 2017-02-14

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