# 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.

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You should assume that $|f^{-1}(w)-P|<r.$ – P.. Nov 12 '12 at 7:06

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 $U$.

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After some thought, it makes sense after applying the Cauchy Integral Formula to the inverse function and then making the substitution $f^{-1}(s) = t$ where t is taken along the path given by $f^{-1}(s)$ where s is along $\partial D(P,r)$.

The problem is being able to deform that curve into a circle which goes around $f^{-1}(w)$. This can be done if we can show the path only goes around $f^{-1}(w)$ once.

I still have no idea how the argument principle is involved

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It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle. – user27126 Nov 15 '12 at 17:59
I would but I still don't have a proof of this, can you direct me to one? – Mathstudent Nov 15 '12 at 18:15
What's wrong with the other answer? – user27126 Nov 15 '12 at 18:57
The other answer is a hint, not a proof of the the above, which I am still confused on. – Mathstudent Nov 19 '12 at 2:17
OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) \neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now? – user27126 Nov 19 '12 at 6:53

1. $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.
1. 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: $$f^{-1}(w) = \frac{1}{2\pi i }\oint_{\partial D(P,r)}\frac{zf'(z)}{f(z)-w}dz$$