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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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The complex conjugate of an element of the unit circle is its inverse. –  Qiaochu Yuan Feb 24 '11 at 23:20
    
@Qiaochu Yuan: You might want to be more specific than "inverse"—e.g. multiplicative inverse or reciprocal. –  Isaac Feb 24 '11 at 23:29
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1 Answer

up vote 7 down vote accepted

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. –  I Love Cake Feb 28 '11 at 4:13
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