Proving theorem connecting the inverse of a holomorphic function to a contour integral of the function.
I am asked to prove this theorem:
If $f:U \rightarrow C$ is holomorphic in $U$ and invertible, $P\in U$ and if $D(P,r)$ is a sufficently small disc about P, then
$$f^{-1}(w) = \frac{1}{2\pi i} \oint_{\partial D(P,r)}{\frac{sf'(s)}{f(s)-w}}ds$$
The book says to "imitate the proof of the argument principle" but I am not seeing the connection.
Hint: Since $f$ is holomorphic and invertible, for each $w\in f(D(P,r))$, $f(z)-w$ has a unique zero $f^{-1}(w):=z_0$ in $U$; moreover, $z_0\in D(P,r)$. Therefore, $f(z)=w+(z-z_0)h(z)$ on $U$, where $h$ is holomorphic and has no zero on $U$.
I had a similar problem:
- $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.
My solution:
- Apply Cauchy Integral formula: Since $f^{-1}(w)$ is analytic by the inverse function theorem, we can say:
$$f^{-1}(w) = \frac{1}{2\pi i }\oint_{f(\partial D(P,r))}\frac{f^{-1}(u)}{u-w}du$$ 2. Using Substitution: $u=f(z)$ for $z$ on $\partial D(P,r) \implies du = f'(z)dz$
3.Rewrite: $$f^{-1}(w) = \frac{1}{2\pi i }\oint_{\partial D(P,r)}\frac{zf'(z)}{f(z)-w}dz$$