Proving $g(\omega)=\frac{1}{2\pi i}\int_{\gamma}\frac{zf'(z)}{f(z)-\omega}\, dz$ where $g$ is the inverse of $f$ I have the following exercise:

Let $G$ be an open subset of $\mathbb{C}$ and let $f$ be a one to one
  function in $H(G)$ such that $f'(z)\neq0$ for all $z\in G$.
For each $\omega\in f(G)$ let $g(\omega)$ denote the unique complex
  number $z\in G$ for which $f(z)=\omega$.
Suppose that the closed disk $\overline{D(z_{o},r)}\subseteq G$ and
  that $\gamma:[0,2\pi]\to G$ is the curve given by
  $\gamma(t)=z_{0}+re^{it}$.
Prove, with the help of the residue theorem, that, for every
  $\omega\in f(D(z_{0},r))$ $$g(\omega)=\frac{1}{2\pi
 i}\int_{\gamma}\frac{zf'(z)}{f(z)-\omega}\, dz$$
Your solution should also explain why this integral is well defined.

I will divide my question into two parts: 


*

*I have a solution for the case that $g(\omega)\neq0$, but there
is an argument made in it that I can't totally justify:



Consider the function  $$ \frac{zf'(z)}{f(z)-\omega} $$
it have the singular point when  $$ f(z)=\omega $$
since $f$ is $1-1$ and $\omega$ is in the image of $f$, there exist a
  unique point $z_{1}$ s.t  $$ f(z_{1})=\omega $$
Since $f'(z)\neq0$ we have it that  $$ h(z):=f(z)-\omega $$
have a zero of order $1$. assume $z_{1}$ maps to $\omega$
$z_{1}f'(z_{1})\neq0$ and $f(z_{1})-\omega=0$, $(f-\omega)'=f'\neq0$
  hence  $$ Res\frac{zf'(z)}{f(z)-\omega} $$
at $z=z_{1}$ is  $$ \frac{z_{1}f'(z_{1})}{(f(z)-\omega)'_{z_{1}}}=z_{1}
 $$
and by the residue theorem  $$ \frac{1}{2\pi
 i}\int_{\gamma}\frac{zf'(z)}{f(z)-\omega}=\frac{1}{2\pi i}2\pi
 iz_{1}=z_{1}=g(\omega) $$

The part that I can't justify is why $h$ have a zero of order $1$
? I think its because $f$ is $1-1$, but I'm not sure how to use
it
2) I don't think it holds for when $g(\omega)=0$, since then $$z_{1}f'(z_{1})=0\cdot f'(0)=0$$
and so I can't use the theorem I used.
Can someone please help me with this case ?
Added later:
I am referring to the following theorem, that can be found at page
$253$ of the book Complex Variables and Applications by Brown and
Churchill:

Let two functions $p,q$ be holomorphic at a point $z_{0}$. If
  $p(z_{0})\neq0,q(z_{0})=0$ and $q'(z_{0})\neq0$ then
  $$Res_{z=z_{0}}\frac{p(z)}{q(z)}=\frac{p(z_{0})}{q'(z_{0})}$$

 A: About 1: notice that if a function $f(z)$ has a zero of order $n>1$ then, $f'(z)=0$, which is excluded in your case.
A: If $F(z) = \displaystyle \frac{A(z)}{B(z)}$ where $A(z)$ and $B(z)$ are analytic, and $B(z)$ has a simple zero at $z = z_1$ (so $B'(z_1) \ne 0$), then the quoted Churchill-Brown's theorem says:
$$Res( F(z); \text{ }z = z_1) = \displaystyle \frac{A(z_1)}{B'(z_1)}$$
This holds even when $A(z_1) = 0$.   In the case when $A(z_1) = 0$, then the function $F(z)$ is either analytic at $z = z_0$ or has a removable discontinuity there. In this case, $Res( F(z) ; \text{ }z = z_1) = 0$.
Either way you get the result that you want: $Res( F(z) ; \text{ }z = z_1) = g(w)$.
A good way to see why $Res( F(z); \text{ }z = z_1) = \displaystyle \frac{A(z_1)}{B'(z_1)}$
is true (as long as $B'(z_1) \ne 0$) near an isolated simple pole or removable singularity is to write out $A(z)$ and $B(z)$ as power series near the singularity, and factor out $(z - z_1)$ out of the denominator power series and see what's left over.
The Bak/Newman book Complex Analysis has a great exposition of Laurent and Power series, and residues.
