# 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:

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.

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

• Why do you think $z_{1}f'(z_{1})=0\cdot f'(0)=0$ prevents your argument from working? One thing you can do to see it works anyway is by expressing the numerator and denominator as power series near $z_1$, then factoring out $f'(z_1)(z - z_1)$ out of the denominator, so as to factor out $\displaystyle \frac{1}{f'(z_1)(z - z_1)}$ out of the entire expression. The resulting fraction of power series is analytic at $z_0$, so has its own Taylor series there. What is the Laurent series you get when you reincorporate $\displaystyle \frac{1}{f'(z_1)(z - z_1)}$ back in? Commented Jul 1, 2013 at 2:40
• @bryanj - please see my edit. I am not sure that I understand your idea, could you give some more explicit details please ? Commented Jul 1, 2013 at 10:38
• I guess what I was getting at is that if $z_1 = 0$, so that $p(z_0) = 0$ (in Churchill's Theorem), but all the other hypotheses are satisfied, then $\displaystyle \frac{p(z)}{q(z)}$ does not have a singularity at $z_0$, so the residue is zero and you still get back $0 = z_1 = g(w)$ when you integrate. Commented Jul 1, 2013 at 11:14
• @bryanj - I think you are right, you might want to post an answer so I can accept. I have to treat this as a special case though, right ? Commented Jul 1, 2013 at 11:16

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.

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.