# $f$ is analytic, and maps the unit disk to itself; $f(0)=0$. Prove $|f(z)|\leq |z|$ for $z\in D$; $|f'(0)|\leq 1$

I am having difficulties with the following problem:

$\bf Given$: $f$ is an analytic map from unit disk $D$ to itself and: $f(0)=0$. $\bf To \; prove:$ $|f(z)|\leq |z|$ for $z\in D$; and: $|f'(0)|\leq 1$

What I thought is: $$|f(z)|\leq 1,$$ because $f$ maps to unit disk.

This is apparently wrong. But how then should I approach this problem?

• The expression $f(z) \leq |z|$ is not meaningful: $\leq$ is not defined for complex numbers. Do you mean $|f(z)|$? – Ulrik Jan 19 '14 at 18:44
• @Svinepels Thank you, yes indeed. I changed it. – user104662 Jan 19 '14 at 18:46
• Why do you need to prove $f(0)=0$? You are given that. – Robert Israel Jan 19 '14 at 18:51
• – user110822 Jan 19 '14 at 18:54
• @user104662 $\lim\limits_{h \to 0} \dfrac{f(h)}{h}=\lim\limits_{h \to 0} \dfrac{f(h)-f(0)}{h-0}=f'(0)$ – user110822 Jan 19 '14 at 19:01

Hint: consider $g(z) = f(z)/z$.