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I want to compute $\int_{|z|=R} \frac{|dz|}{|z-a|^4}$ when $R>0, |a|\neq R$. I did so by parametrizing the circle, but this way required many pages of tedious calculations.

Do you have a more straightforward way to do it?

Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.

Thanks

perlman
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2 Answers2

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Try This

$f^n(z)$ = $\frac {n!}{2(pi)n}$ $\int$ $\frac {f(w) dw}{(w-z)^{n+1}}$

Math Model
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  • Can you make it a bit more clear? Specifically, how do I treat the $|dz|$? – perlman Oct 24 '17 at 05:47
  • I just assumed the absolute value sign was meant to be around the function. I don't think I've specifically dealt with a case you're talking about, because I tried googling it, and |dz| turns out to be some kind of arc length. – Math Model Oct 24 '17 at 05:57
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After parametrising $z= R e^{i\theta}$ and after a lot of simplifications you will get $$I = \int_{0}^{2\pi} d\theta \frac{R}{[(R\cos\theta-Re(a))^2+(R\sin\theta-Im(a))^2]^2_{}}.$$ To proceed further : go back to complex variable $Z = e^{i\theta}$, rewrite it in terms of $\cos\theta=\frac{1}{2}(z+\frac{1}{z})$ and $\sin\theta=\frac{1}{2i}(z-\frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.

Sunyam
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