I need some help regarding why the contour integral $$\int_{\gamma} \frac{f(w)}{w-\frac{1}{\bar{z}}} \mathrm dw$$ is equal to zero, where $\gamma$ is the unit circle and $f$ is holomorphic on the unit disk and continuous on its closure. I think I have to use Cauchy's Theorem, but can't understand why the integrand in this case is holomorphic and why, for example, the integrand in Cauchy's Formula (i.e. $\frac{f(w)}{w-z}$) being very similar, is not.

Thank you all for any clues you can give me

  • $\begingroup$ Where is $z$ located relative to $\gamma$? $\endgroup$ – md2perpe Jan 30 at 16:58
  • $\begingroup$ z in on the open unit disk, so that $\frac{1}{\bar{z}}$ is outside the unit circle $\gamma$. The thing is that I don't know how's that related to the function being holomorphic, since $w$ is in the unit circle in both of the cases I mentioned in my question $\endgroup$ – xan32 Jan 31 at 0:03

When $\zeta$ is inside $\gamma$ then $1/(w-\zeta)$ is not holomorphic inside $\gamma$ since it has a pole at $w=\zeta$. When $\zeta$ is outside $\gamma$ then $1/(w-\zeta)$ has no pole inside $\gamma$ and is therefore holomorphic there. Thus, if $z$ is inside $\gamma$ then $1/(w-z)$ is not holomorphic, but $1/(w-1/\bar z)$ is. Therefore the integral in your question vanishes.


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