I am revising for an exam, and the following (first) step in the proof of Schwarz Integral Formula is presented without derivation. I don't understand it, can somebody help?
"$\Gamma$ is a circle (perifery) $|z-z_0|=R$, $f(z)$ is analytic inside and continuous on the boundary of the circle $C(R,z_0)$.
Applying Cauchy Integral Formula, $f(z)=\frac{1}{2\pi i}\int_{\Gamma}{\frac{f(t)dt}{t-z}}$, $z\in C(R,z_0)$.
now is the part i don't understand:
Then $0=\int_{\Gamma}{\frac{f(t)dt}{t-z^{*}}}$, where $z^{*}=z_0+\frac{R^2}{\bar{z}-\bar{z_0}}$. (Why?)"
Best Answer
$z^{*}$ is outside the circle of radius $R$ around $z_0$ if $z$ is inside it. This makes the integrand analytic so it is $0$ by Cauchy's Theorem.