Classify the singularities of $$\frac {\sin(2\pi z)}{z^3(2z-1)}.$$
So I know that the singularities are at $ z = 0 $ and $ z = 1/2 $. Since $$ \sin(2\pi z) = 2\pi z - \frac{(2\pi z)^3}{3!} + \ldots,$$ then I believe that $z = 0$ is a pole of order $2$, but for $z = \frac{1}{2} $ we have that $\frac {\sin(2\pi z)}{z^3(2z-1)}$ is analytic, does this make $\frac {1}{2}$ a removable singularity?