Complex integral $\oint_{|z|=1} \frac{z^2}{\sin^3 z \cos z} dz$

Question

Hello I am trying to evaluate: $$I=\oint_{|z|=1} \frac{z^2}{\sin^3 z \cos z} dz$$ My try was to compute the residue at infinity since $I=2\pi i Res(f,\infty)$ . Now since $$\sin^3 z= \frac{3}{4}\sin z -\frac{1}{4}\sin(3z)$$ Expanding into series gives: $$\sin^3 z =\frac{1}{4}\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!}(3-3^{2n+1})$$ therefore with the series of $\cos z$ we have :$$\frac{z^2}{\sin^3 z \cos z}=\frac{z^2}{\frac{3}{4}\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!}(1-3^{2n})\sum_{n=0}^{\infty}(-1)^n \frac{z^{2n}}{(2n)!}}$$ Now I thought to do a cauchy product with the two series, but here is the problem, I dont have equal powers, and since one is odd and another even I dont know how to equate those power in order to be able to do a Cauchy product, could you help me? Or maybe is there a easier way to solve this integral?

Answer 1

The residue at $z=0$ is $1$ because the denominator is $z^3$ plus higher order terms in $z$. So if this is the only pole in the region, you are through.

Now look for solutions to $\sin z = 0$ with $0 < |z| \leq 1$ -- They don't exist (the zeros of sine are only at $n\pi$). Similarly, the zeros of cosine are only at $(n+\frac12) \pi$. So there is only the one pole in this region.