Cauchy's Theorem and maximum modulus principle Suppose $f\in H(\Omega), \Gamma$ is a cycle in $\Omega$ such that $Ind_{\Gamma}(\alpha)=0$,for all$\alpha \notin \Omega$，$|f(\zeta)|\leq 1$ for every $\zeta \in \Gamma$,
and $Ind_{\Gamma}(z) \neq 0 $.Prove that $|f(z)|\leq 1$
Cauchy's Theorem implies:
$$ |f(z)\cdot Ind_{\Gamma}(z)|=\frac{1}{2\pi }|\int_{\Gamma}\frac{f(\omega)}{\omega-z}dz|$$
But I can't estimate the integral.
I will appreciate your help.

Oh,I found the answer.
$Ind_{\Gamma}(\alpha)=0$,for all $\alpha \notin \Omega$ implies:
The union of $\Gamma$ and the components of the complement of E which $Ind_{\Gamma}\neq 0 $ is 
$ \subset \Omega$.This is the point of the question. Then the boundary of the component contained z is $\subset \Gamma$ and maximum modulus principle implies $|f(z)|\leq 1$.
There is an example about the importance of $Ind_{\Gamma}(\alpha)=0$,for all $\alpha \notin \Omega$:
Suppose $\Omega={1/2<|z|<2}$,$\Gamma$ is the positively oriented circle with center at 0 and radius $1$,$f=1/z$.
If we follow the answer above,we can find the boundary of the union contained $|z|=1/2$.
