Inverse of a univalent analytic function

Question

Let $f: U \rightarrow \Bbb C$ be a univalent analytic function, and let $\gamma$ be a Jordan contour in $U$ with the property that the inside $D$ of the Jordan curve $ |\gamma |$ is contained in $U$. Demonstrate that for every point $w$ of the domain $D^ {\prime}=f(D)$ the value of the inverse of $f$ at $w$ is given by the formula

$f^{-1}(w) =\frac {1}{2 \pi i} \int\limits_\gamma \frac {zf^{\prime}(z)dz}{f(z)-w}$

My work: Fix $w \in f(D)$ and let $f(z_0)=w, a \in D$. By the definition of univalent analytic function, let $h(z) =f(z)-f(a)=f(z)-w$ $$g(z)=\frac{zf'(z)}{f(z)-w}.$$ The only singularity of $g$ is at $a=f^{-1}(w)$, and it's a simple pole. The residue there is $$\lim_{z\to a}\left(zf'(z)\frac{z-a}{f(z)-f(a)}\right).$$ i.e The residue of $Res(a, \frac{zf'(z)}{f(z)-w})=a$. Thus by residue theorem, $ \int\limits_\gamma \frac {zf^{\prime}(z)dz}{f(z)-w}=2 \pi i (a)$ Since $f^{-\prime} (w)=f^{-\prime} (f(a))=a$, we obtain $f^{-1}(w) =a=\frac {1}{2 \pi i} \int\limits_\gamma \frac {zf^{\prime}(z)dz}{f(z)-w}$ Did I miss anything? I appreciate your kind help. Thank you!

Answer 1

Fix $w$ and let $$g(z)=\frac{zf'(z)}{f(z)-w}.$$ The only singularity of $g$ is at $a=f^{-1}(w)$, and it's a simple pole. The residue there is $$\lim_{z\to a}\left(zf'(z)\frac{z-a}{f(z)-f(a)}\right).$$ Can you take it from here?

Answer 2

Hint: By the Generalized Argument Principle, the integral $$ \frac1{2\pi i}\int_\gamma\frac{zf'(z)}{f(z)-w}\,dz $$ equals $g(a)$ where $g$ is the identity function $z\mapsto z$ and $a$ is the zero of $f(z)-w$ inside $\gamma$.