simple pole of a complex function

Question

Given $$f(z)=\dfrac{\tanh (2z)}{z\sin(\pi z)}$$ Are $z=\pm 1$ simple poles of $f(z)$? It appears so as $\sin(n\pi)=0$ for integer n. But the integral along the circle $|z|=\pi/2$ of this function is given as $4i$. Did I go wrong?

Answer 1

Yes, they are poles of order $1$ but the residues at these points cancel out. It has a simple pole at $0$. The residue is $\frac 2 {\pi}$ so the integral is $2\pi i \frac 2 {\pi}=4i$.

Answer 2

$$f(z)=\dfrac{\tanh (2z)}{z\sin(\pi z)}=g(z)h(z)$$ where $g(z)=\frac{\tanh (2z)}{z}$ and $h(z)=\frac{1}{sin(\pi z)}$.

Note that $g(z)$ had removable singularity at $z=0$ while $h(z)$ has simple pole there, so $f(z)$ also has simple pole at $z=0$ in addition to the simple poles at $z=\pm 1$