2
$\begingroup$

I am trying to compute the following integral$\int_{\partial D\left(0,1\right)}\frac{\overline{z}}{z-w}dz$ where $\partial D\left(0,1\right)$ is the boundray of the unit disk in $\mathbb{C}$ . This is what I have so far, but I am not sure what to do next:$\int_{\partial D\left(0,1\right)}\frac{\overline{z}}{z-w}dz=\int_{0}^{2\pi}\frac{e^{-it}}{e^{it}-w}d\left(e^{it}\right)=\int_{0}^{2\pi}\frac{i}{e^{it}-w}dt.$ Any help would be apprecaited.

  • 1
    I suppose |w|<1 ?2012-10-27

2 Answers 2

4

I get a slightly different answer: $\int_{\partial D(0,1)}\frac{\bar{z}}{z-w}dz=\int_{\partial D(0,1)}\frac{\left|z\right|^{2}}{z\left(z-w\right)}dz=-\frac{1}{w}\int_{\partial D(0,1)}\left(\frac{1}{z}-\frac{1}{z-w}\right)dz$ since $|z|=1$ on the unit disk Hence, depending on whether $w$ lies within the unit disk or not the integral is 0 or $-\frac{2\pi\bar{w}}{|w|^2}i$ respectively.

EDIT Obviously the case $|w|=0$ must be treated separately here

  • 0
    Just for comparison sake with the previous answer (now deleted)2012-10-27
0

As it stands the integral $I(w)$ is undefined when $|w|=1$.

On $\partial D$ we have $z\bar z=1$ and therefore $\bar z={1\over z}$. So $I(0)=\int_{\partial D}{\bar z \over z}\ dz=\int_{\partial D}{1\over z^2}\ dz=0\ .$ When $w\ne0$ then $\eqalign{\int_{\partial D}{\bar z\over z-w}\ dz&=\int_{\partial D}{1\over z(z-w)}\ dz = {1\over w}\int_{\partial D}\Bigl({1\over z-w}-{1\over z}\Bigr)\ dz\cr &={2\pi i\over w}\bigl(1_{|w|<1} -1\bigr)\ .\cr}$ Therefore $I(w)=\cases{0&$(|w|<1)$\cr {\rm undefined}&$(|w|=1)$\cr -{2\pi i\over w}&$(|w|>1)$ .\cr}$