Prove that f is one-to-one on D

Question

Let $f$ be an analytic function on a disc $D$ whose center is the point $z_0$.
Assume that $|f'(z)-f'(z_0 )|<|f'(z_0)| $ on D.
Prove that $f$ is one-to-one on D.

Answer 1

Square each side of the inequality to get $$|f'(z)|^2 + |f'(z_0)|^2 -2\operatorname{Re}(f'(z)\overline{f'(z_0)})<|f'(z_0)|^2.$$ This implies that $\operatorname{Re}(f'(z)\overline{f'(z_0)})>0$ for all $z \in D$. Now, if $z,w \in D$ and $z \neq w$, then $$\operatorname{Re}\frac{\overline{f'(z_0)}(f(w)-f(z))}{w-z} = \int_{0}^1\operatorname{Re}\left(\overline{f'(z_0)}(f'(z+(w-z)t)\right) dt$$ which is easily seen to be strictly positive since $\operatorname{Re}(f'(\zeta)\overline{f'(z_0)})>0$ for all $\zeta \in D$. Thus $f(z)\neq f(w)$.