Suppose $f$ is analytic in a convex region $D$ and $|f'|\le 1$ throughout $D$. Prove that $f$ is a contraction, that is, show that $|f(b)-f(a)|\le|b-a|$ for all $a,b$ in $D$.
Since $f$ is analytic, it is differentiable in $D$. This seems like we would use the mean-value theorem, namely there exists $c$ such that $f'(c)=(f(b)-f(a))/(b-a)$. Then we take the absolute value of both sides and $|f'|<1$. So $|f(b)-f(a)|<|b-a|$, but I'm not sure if I can use the Mean Value Theorem applies, since we do not know if $f$ is real-valued. Also I haven't used that the region is convex. Any help would be appreciated. Thanks!