Estimate when $f$ is locally one to one

Question

Suppose $f$ is analytic in $|z|\leq 1$ and $|f(z)|<1$, $f(0)=0$ and $f '(0)=a\neq 0$. Show that there exists a disc of radius $\rho$ s.t for any $z_1,z_2$ in the disc, $f(z_1)=f(z_2)$ implies that $z_1=z_2$. I know this has to do with the map $(z-\alpha)/(1-\overline{\alpha}z)$. with $|\alpha|<1$

Answer 1

Hint: Choose $r>0$ such that $|f'(z)-f'(0)| < |f'(0)|/2 , z\in D(0,r).$ For $z,w\in D(0,r)$ we have

$$f(w)-f(z) = \int_{[z,w]} f'(\zeta)\ d\zeta = \int_{[z,w]} (f'(\zeta)-f'(0))\ du +f'(0)(w-z).$$