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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.

Gerry Myerson
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Mike
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4 Answers4

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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 $U$.

23rd
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I had a similar problem:

  1. $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.

My solution:

  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: $$f^{-1}(w) = \frac{1}{2\pi i }\oint_{\partial D(P,r)}\frac{zf'(z)}{f(z)-w}dz$$

J_Lopez8
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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

Martin Argerami
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Mathstudent
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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. –  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? –  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 02:17
  • 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? –  Nov 19 '12 at 06:53
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The answer by J_Lopez8 gives a motivation (and a proof) for the "mysterious" formula. But knowing the formula, you can prove it using residue theory. As $f'\ne 0$ in a nhood of $P$, for $r$ small enough the integrand has at most a simple pole in $f^{−1}(w)$ with residue $$ {\rm Res}_{s=f^{-1}(w)}{\frac{sf'(s)}{f(s) - w}} = \lim_{s\to f^{-1}(w)}(s - f^{-1}(w))\frac{sf'(s)}{f(s) - w} = \lim_{s\to f^{-1}(w)}\frac{{sf'(s)}}{\frac{f(s) - f(f^{-1}(w))}{s - f^{-1}(w)}} = \frac{f^{-1}(w)f'(f^{-1}(w))}{f'(f^{-1}(w))} = f^{-1}(w). $$ For an alternative proof (Complex Functions by Goodstein, 8.2.8), use the following theorem:

Let be $f:D\longrightarrow{\Bbb C}$ holomorphic in the simply connected open set $D$, $\gamma\subset D$ a simple closed curve, $f'(z)\ne 0$ for all $z\in\gamma$. If $f$ has a only simple pole (has only a pole and is simple) $z_0$ in the interior of $\gamma$, then $$z_0 = \frac1{2\pi i}\int_\gamma\frac{z f'(z)}{f(z)}\,dz.$$

Martín-Blas Pérez Pinilla
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