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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?2012-07-23
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    The residue theorem?2012-07-23
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    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 1.2012-07-23
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    @TigranHakobyan : Are you using $x$ and $z$ synonymously.2012-07-23
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    @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.2012-07-23
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    @MichaelHardy: surely that simplifies the question even more then? No poles within the region of integration means, as the answer says, that the integral is 0.2012-07-23
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    OK, I've posted an answer. I think it's a bit simpler than any that explicitly tries to calculate the residue.2012-07-23

2 Answers 2

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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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$$\text{If}\,\,f(z)=\frac{\sin z^2}{\sin^2z}\,\,,\,\text{then}\,\,Res_{z=0}(f)=\lim_{z\to0}\frac{d}{dz}\left(z^2\frac{\sin z^2}{\sin^2z}\right)=0$$

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 $\,\displaystyle{\frac{1}{1-z}\,\,,\,|z|<1}\,$