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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. $\endgroup$ – TZakrevskiy Sep 30 '13 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 '13 at 2:44
  • $\begingroup$ Partial fraction decomposition? You do not want to do it by parametrizing explicitly. $\endgroup$ – Ted Shifrin Sep 30 '13 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 '13 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: $z=2e^{i\theta}$, $dz=\dots\,d\theta$. $\endgroup$ – Ted Shifrin Sep 30 '13 at 3:18
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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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