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Compute $\int_{|z| = r} \frac{|dz|}{|z-a|^2}$ where $|a| \neq r$

I know that $|dz| = -ir\frac{dz}z$ and that $z\overline{z} = r^2$. So I get

$\int_{|z| = r} \frac{|dz|}{|z-a|^2} = -i\int_{|z| = r} \frac{rdz}{(z-a)(\overline{z} - \overline{a})} $

But I don't know how to proceed from here

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1 Answer 1

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The denominator of your integrand is missing a factor of $z$. The denominator should be $z(z-\alpha)(\bar{z}-\bar{\alpha})$. The idea now is to multiply this out and replace $z\bar{z}$ with $r^2$. After some multiplication and factoring, the denominator of your integrand becomes $(r^2-\bar{\alpha}z)(\alpha-z)$.

Thus, your integral takes the form: $$\int_{|z|=r}\frac{-ir}{(r^2-\bar{\alpha}z)(\alpha-z)} = \int_{|z|=r}\frac{ir}{(r^2-\bar{\alpha}z)(z-\alpha)} = \int_{|z|=r}\frac{ir}{(r^2-\bar{\alpha}z)}\cdot\frac{1}{(z-\alpha)}$$

The answer now depends on whether $|\alpha| > r$ or $|\alpha| < r$. In the latter case, use Cauchy's integral formula.

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