If $f$ is a holomorphic function in $D(0, p)$ and let $\gamma(t) = re^{it}, 0≤t≤2\pi$, with $0 < r < p$. If $b$ is not in $D(0,p)$, and $z-b$ is not holomorphic then $\int_{\gamma}\dfrac{f(z)}{z-b} dz = 0$
I don't understand why that is true
Integration of non holomorphic function
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complex-analysis
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1How can $z-b$ be non holomorphic? Is $b$ a complex constant? – 2017-01-10
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0@ajotatxe sorry for the lack of details, I edited my post – 2017-01-10
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0But, still, the function $z\mapsto z-b$ is holomorphic. (Then the statement is trivially true, but I guess that this is not what you mean). – 2017-01-10
1 Answers
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It seems that you have copied the statement from a book, lecture notes or something.
It seems a typo. It is true that $\int_\gamma\frac{f(z)}{z-b}dz=0$. But the reason is that $f$ is holomorphic in a disc that contains the image of $\gamma$ and $z-b$ is not zero in this disc. Hence $f(z)/(z-b)$ is holomorphic, too.