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Let $f$ be a holomorphic function on the disk

$$ D_{r_0} =\{z\in\ {C} : |z|<R_0 \} $$ centered at the origin and of radius $R_0$. $$$$ Prove that whenever $0<R<R_0$ and $|z|<R$, then $$f(z) =\frac{1}{2π}\ \int_{0}^{2\pi} f(Re^{i\phi}) Re \bigl ( \frac {Re^{i\phi}+z}{Re^{i\phi}-z} \bigr) d \phi. $$ and the lecture note start with below

$$ f(z) =\frac{1}{2πi}\ \int_{|ζ|=R} \frac{f(ζ)}{ζ−z}\ dζ. $$

$$ 0 =\frac{1}{2πi}\ \int_{|ζ|=R} \frac{f(ζ)}{ ζ−\frac{R^2}{\bar{z}} }\ dζ. $$

My lecture note says that the second equation holds by Cauchy Theorem. But I don't know why the second equation is equal to zero.$\frac{R^2}{\bar{z}}$ could be on the disk which means ${|ζ|=R}$. Am I wrong?.

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Where does $D_{R_0}$ appear, in your lecture notes? It seems you never use it. –  Siminore Oct 4 '12 at 9:49
    
I don't exactly see how the term $\frac{R^2}{\bar{z}}$ comes up. More specifically, is the second equation supposed to hold for any choice of $z\in\mathbb{C}$? –  Giovanni De Gaetano Oct 4 '12 at 9:50

1 Answer 1

up vote 1 down vote accepted

The second equation hods for $|z|<R$. I assume that in the third equation we also have $|z|<R$. Then $|R^2/\bar z|>R$ and the function $$ \frac{f(z)}{z-R^2/\bar z} $$ is holomorphic on an open set containing $\{|z|\le R\}$. Cauchy's Theorem implies that the integral is zero.

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Thank you very much sir. –  HCCHUNG Oct 4 '12 at 10:39

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