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So the problem was to integrate $\overline{z}$ over the half circle of radius 2, $ z = 2e^{i \theta}$ from $z = -2i$ to $z = 2i$. After going through the math, they say this integral is equal to the integral over the curve of $\frac{dz}{z}$ which is equal to $\pi i $. I don't follow how this equivalence was established. I hope I provided enough detail, still learning how to use MathJax. Thanks!

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    @Qiaochu Yuan: You might want to be more specific than "inverse"—e.g. multiplicative inverse or reciprocal.2011-02-24

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For $z$ on this half circle $z\overline{z}=|z|^2=4$ so that $\overline{z}=\frac{4}{z}$. Let $C$ be the contour in question. Then we have $\int_C \overline{z}=\int_C \frac{4}{z}dz=\int_{-\pi/2}^{\pi/2}\frac{4}{2e^{i\theta}}d(2e^{i\theta})=\int_{-\pi/2}^{\pi/2}4id\theta=4\pi i$

I have an extra factor of $4$. Are you sure that is not correct?

Hope that helps,

Alternative:

Just put in your parametrization directly, that is $z=2e^{i\theta}$. Then we get $\int_C \overline{z}dz=\int_{-\pi/2}^{\pi/2}2e^{-i\theta}d(2e^{i\theta})=\int_{-\pi/2}^{\pi/2}4id\theta=4\pi i$

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    For some reason I thought this was some sort of general result, and not to do so much with the original problem :\ Thanks a lot.2011-02-28