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Consider $z \in \mathbb{C}$ and

$$\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$$

How would we integrate this?

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There are poles within the region enclosed by the contour. What theorem should be screaming at you right now? –  Kris Jul 23 '12 at 20:26
    
The residue theorem? –  Chris Jul 23 '12 at 20:29
    
write the denominator like this: $z^2(\frac{sin(z)}{z})^2$ and then use the residue theorem. Note that we can assume that $\frac{sin(z)}{z}$ function is holomorph, if we set its value at 0 to be 1. –  Tigran Hakobyan Jul 23 '12 at 20:39
    
@TigranHakobyan : Are you using $x$ and $z$ synonymously. –  Michael Hardy Jul 23 '12 at 20:40
    
@Kris : Where do you find these poles? Inside the circle, the only place where the denominator is $0$ is at $z=0$, and that's a removable singularity. –  Michael Hardy Jul 23 '12 at 20:41
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2 Answers

up vote 5 down vote accepted

The sine is an entire function, i.e. has no sigularities at finite points. So the only thing that could cause bad behavior is zeros in the denominator.

The only place inside the circle $|z|=1$ where the denominator is $0$ is $z=0$. Now notice that $$ \lim_{z\to0} \frac{\sin(z^2)}{(\sin z)^2} = 1, $$ so there's no pole there. It's a removable singularity. So you're integrating around a circle a function that has no singularities inside the circle, so you get $0$.

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Thank you very much! –  Chris Jul 23 '12 at 21:09
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$$\text{If}\,\,f(z)=\frac{\sin z^2}{\sin^2z}\,\,,\,\text{then}\,\,Res_{z=0}(f)=\lim_{z\to0}\frac{d}{dz}\left(z^2\frac{\sin z^2}{\sin^2z}\right)=0$$

So the singularity at $\,z=0\,$ is in fact a removable one and thus the integral equals zero.

We can also use power series in a tiny neighbourhood of zero: $$\frac{\sin z^2}{\sin^2z}=\frac{z^2-\frac{z^6}{3!}+\cdots}{\left(z-\frac{z^3}{3!}+\cdots\right)^2}=\frac{z^2\left(1-\frac{z^4}{3!}+\cdots\right)}{z^2\left(1-\frac{z^2}{3}+\cdots\right)}=\left(1-\frac{z^4}{3!}+\cdots\right)\left(1+\frac{z^4}{3}+\cdots\right)$$ the rightmost parentheses being the power series for $\,\displaystyle{\frac{1}{1-z}\,\,,\,|z|<1}\,$

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