Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ Consider $z \in \mathbb{C}$ and
$$\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$$
How would we integrate this?
 A: 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 $0$.
A: $$\text{If}\,\,f(z)=\frac{\sin z^2}{\sin^2z}\,\,,\,\text{then}\,\,Res_{z=0}(f)=\lim_{z\to0}\frac{d}{dz}\left(z^2\frac{\sin z^2}{\sin^2z}\right)=0$$
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 $\,\displaystyle{\frac{1}{1-z}\,\,,\,|z|<1}\,$
