Theorem connecting the inverse of a holomorphic function to a contour integral

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SUMMARY

The discussion centers on the proof of a theorem connecting the inverse of a holomorphic function to a contour integral. Specifically, it establishes that if \( f: U \to \mathbb{C} \) is holomorphic and invertible in a neighborhood \( U \), then the inverse function can be expressed using Cauchy's integral formula. The formula states that for a point \( P \in U \) and a sufficiently small disc \( D(P,r) \), the inverse \( f^{-1}(w) \) is given by \( f^{-1}(w) = \frac{1}{2\pi i} \oint_{\partial D(P,r)} \frac{sf'(s)}{f(s)-w} ds \). This relationship is crucial for understanding the behavior of holomorphic functions and their inverses.

PREREQUISITES
  • Understanding of holomorphic functions
  • Familiarity with contour integrals
  • Knowledge of Cauchy's integral formula
  • Basic complex analysis concepts
NEXT STEPS
  • Study the implications of Cauchy's integral formula in complex analysis
  • Explore the properties of holomorphic functions and their inverses
  • Learn about the application of contour integrals in proving theorems
  • Investigate the relationship between differentiability and holomorphicity in complex functions
USEFUL FOR

Mathematicians, students of complex analysis, and researchers interested in the properties of holomorphic functions and their inverses will benefit from this discussion.

IDontKnow1
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I tried posting this at stack exchange but it never got the question answered. I want to prove this:

If f:U→C is holomorphic in U and invertible, P∈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
 
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That is just Cauchy's formula.

f^{-1}(w)=\frac{1}{2\pi\imath}\oint_{\partial D(P,r)} \frac{s}{f(s)-w}d(f(s))
 

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