Classifying singularities of $\frac{\sin(\pi z)}{z^4+1}$ If $f(z)=\frac{\sin(\pi z)}{z^4+1}$, we have four roots of unity, which are isolated singularities of $f$: $$z=-(-1)^{1/4},z=(-1)^{1/4}, z=-(-1)^{3/4}, z=(-1)^{3/4}.$$
Do we need to find the Laurent series about all four of the singularities in order to classify them, or is there a better method?

No need to compute the Laurent series. All the singularities are zeroes of order $1$ of the denominator, and the numerator does not vanish at them. This implies that they are poles of order one. You can see it as follows. Call $z_i$, $1\le i\le4$ the singularities. Then
$$
f(z)=\frac{1}{z-z_1}\,\frac{\sin(\pi\,z)}{(z-z_2)(z-z_3)(z-z_4)},
$$
and
$$
\lim_{z\to z_1}(z-z_1)f(z)=\frac{\sin(\pi\,z_1)}{(z_1-z_2)(z_1-z_3)(z_1-z_4)}\ne0.
$$
Similarly for the other singularities.
