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Compute $\int_{|z|=2}\dfrac{1}{z^2-1}dz$ where the circle $|z|=2$ is oriented counterclockwise.

I want to use the formula $\int f(z)dz=\int_a^b f(z(t))z'(t)dt$. So I parametrize the curve by $z(t)=2(\cos t+i\sin t)$ for $0\le t\le 2\pi$.

The integral becomes $\int_0^{2\pi}\dfrac{1}{4(\cos 2t+i\sin 2t)-1}\cdot(-2\sin t+2i\cos t)dt$.

This is a mess. How would I integrate it?

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    $\begingroup$ If I were you I would use Residue theorem to calculate this integral. $\endgroup$ Sep 30, 2013 at 2:32
  • $\begingroup$ @TZakrevskiy Interesting, but in the book I'm reading (Ahlfors), in the section in which the exercise appears, the book hasn't got to Residue theorem yet. Is there another way to do it? $\endgroup$
    – PJ Miller
    Sep 30, 2013 at 2:44
  • $\begingroup$ Partial fraction decomposition? You do not want to do it by parametrizing explicitly. $\endgroup$ Sep 30, 2013 at 2:47
  • $\begingroup$ @TedShifrin Sure, I can do $\dfrac{1}{z^2-1}=\dfrac12\left(\dfrac1{z-1}-\dfrac1{z+1}\right)$ and integrate each part. But what do you mean by not parametrizing explicitly? $\endgroup$
    – PJ Miller
    Sep 30, 2013 at 2:53
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    $\begingroup$ Oh, on second thought, you can easily do the partial fractioned version by parametrizing. Just keep it in complex terms: $z=2e^{i\theta}$, $dz=\dots\,d\theta$. $\endgroup$ Sep 30, 2013 at 3:18

1 Answer 1

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Let $C:|z|=2$

$z(\theta)=2e^{i\theta}$ where $0\leq\theta\leq 2\pi$

$z'(\theta)=2ie^{i\theta}d\theta$

$\int_C \frac{dz}{z^2-1}=\int_0^{2\pi}\frac{2ie^{i\theta}d\theta}{(2e^{i\theta})^2-1}$

let $u=2e^{i\theta}\implies \frac{du}{i}=2e^{i\theta}d\theta$

By partial fractions:

$\int \frac{du}{u^2-1}=\frac{1}{2}[ln(2e^{i\theta}-1)-ln(2e^{i\theta}-1)]$ evaluated at $0$ and $2\pi$

and $2e^{i2\pi}=2=2e^{i0}$ so we have $\frac{1}{2}[ln(2-1)-ln(2-1)-ln(2-1)+ln(2-1)]=0$.

Thus $\int_C \frac{dz}{z^2-1}=0$.

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