0
$\begingroup$

Let $f$ be a holomorphic function on the disk

$$ D_{r_0} =\{z\in\ {C} : |z|

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

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

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?.

  • 0
    Where does $D_{R_0}$ appear, in your lecture notes? It seems you never use it.2012-10-04
  • 0
    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 $z\in\mathbb{C}$?2012-10-04

1 Answers 1

2

The second equation hods for $|z|R$ 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.