Let $f(z)=\frac{1}{(1-z)^2}$ and let $0<R<1$ prove that $\max_{|z|=R}{|f(z)|}=\frac{1}{(1-R)^2}$

Question

Let $f(z)=\frac{1}{(1-z)^2}$ and let $0<R<1$ prove that $\max_{|z|=R}{|f(z)|}=\frac{1}{(1-R)^2}$

i am just starting with suppose $f(z)>\frac{1}{(1-R)^2}$ for some $|z|=R$ but i cant go for further

Answer 1

$$\max_{|z|=R}\frac{1}{|1-z|^2} = \left(\min_{|z|=R}|1-z|^2\right)^{-1} = \frac{1}{|1-R|^2} $$ since the point on the circle $|z|=R$ closest to $z=1$ is pretty obviously $z=R$.