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.
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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 |
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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 $$I(w)=\cases{0&$(|w|<1)$\cr {\rm undefined}&$(|w|=1)$\cr -{2\pi i\over w}&$(|w|>1)$ .\cr}$$ |
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