If $f\in S$ and $|f|\leq M$ in the unit circle, then show that $g(z)=\frac{f(z)}{[1+\frac{e^{i\gamma}}{M}f(z)]^2}$ is univalent. [closed]
The function $h(w)=\frac{w}{(1-w)^2}$ is the Koebe function which is univalent for $|w|<1$ so writing $$-\frac{e^{i\gamma}}{M}g(z)=\frac{-\frac{e^{i\gamma}}{M}f(z)}{[1-(-\frac{e^{i\gamma}}{M}f(z))]^2}$$ and noting that if $w=-\frac{e^{i\gamma}}{M}f(z)$ we have $|w|<1$ by hypothesis (and maximum modulus since $f$ cannot be a constant so $|f(z)|<M$).
Then if $g(z_1)=g(z_2)$ we get by the above $w_1=w_2$ so $z_1=z_2$ since $f$ univalent and we are done!