Past exam of the algebra and analysis I for mathematics and philosophy course of the University of Oxford / 2013 / Analysis / 6 / c / Solution

The complex sine has simple zeros exactly at $z=k\pi$, $k\in \mathbb{Z}$, because $\sin (k\pi )=0$ and $\cos (k\pi )=(-1{)}^k\ne 0$. Hence the denominator has a zero of order three at each of these points and nowhere else.

At $0$, the numerator has order two, so the quotient has a simple pole. Indeed its leading term is $-{\pi }^4/z$. At $\pi$, the numerator has order one, so there is a pole of order two. At $-\pi$, the numerator has order three, cancelling the denominator's order-three zero, so the singularity is removable. Its extension value isAt every remaining $k\pi$, the numerator is nonzero, giving a pole of order three. Thus the complete finite-plane classification isAway from this discrete set the denominator is nonzero, so the quotient is holomorphic.