Contour integral vanishes

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

• Where is $z$ located relative to $\gamma$? – md2perpe Jan 30 at 16:58
• 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 – 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.