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]

Question

I am stuck proving that if $f\in S$, i.e., $f$ is univalent in the unit disc, $f(0)=0$ and $f'(0)=1$, and if $f$ is bounded by a constant $M$ in the unit disc, i.e., $|f(z)|\leq M\ \forall |z|<1$, then the function $g$ defined by $$g(z)=\frac{f(z)}{[1+\frac{e^{i\gamma}}{M}f(z)]^2}$$where $0\leq\gamma<2\pi$, is also univalent in $|z|<1$.

Answer 1

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!