> 
> Suppose $f(a)=0$ for some $|a|<1$. If $f$ is analytic in $D$ containing the unit disk and $|f(z)|<M$ finite for all $z:|z|=1$, show for all $z:|z|<1$, that
>  $$
> |f(z)| \leq M \left|\frac{z-a}{\bar a z - 1}\right|.
> $$
> 
> 
> 

**Attempt**

Since $f(a)=0$,
$$
\frac{1}{2\pi i} \int\_C \frac{f(z) dz}{z} = 0.
$$
So
$$
\left| \int\_C \frac{f(z) dz}{z} \right|
\leq \int\_C \frac{|f(z)|}{|z|}dz
\leq M \int\_C \frac{1}{|z|}dz
= M 2 \pi.
$$
But is there a relationship between $\int\_C \frac{1}{|z|}dz$ and $\left|\frac{z-a}{\bar a z - 1}\right|$? It seems like the integral is just equal to $2\pi$. I need a better bound.

Perhaps we should use the fact that $w = \frac{z-a}{\overline a z - 1}$ maps the closed unit disk in the $z$-plane to the closed unit disk in the $w$-plane, but I don't know if this works with my approach.