Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

2 Answers

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 $U$.

share|improve this answer
add comment

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

share|improve this answer
    
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. –  Sanchez Nov 15 '12 at 17:59
    
I would but I still don't have a proof of this, can you direct me to one? –  Mike Nov 15 '12 at 18:15
    
What's wrong with the other answer? –  Sanchez 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. –  Mike 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? –  Sanchez Nov 19 '12 at 6:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.