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

Question

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:

1. 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 $$Res_{z=z_{0}}\frac{p(z)}{q(z)}=\frac{p(z_{0})}{q'(z_{0})}$$

Answer 1

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.

Answer 2

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.