Let $C$ denote the positively oriented circle $|z|=2$. Evaluate $\int_C \tan zdz$
We are in the residue unit, so I found that there are singularities at $z_k=(\frac{1}{2}+k)\pi$ for $k\in \mathbb{Z}$. Let $p(z)=\sin z$, $q(z)=\cos z$. Then both are analytic at $z_k$ and $q(z_k)=0$ while $q'(z_k)\ne 0$, so $\operatorname{Res}_{z=z_k}f(z)=\frac{p(z_k)}{q'(z_k)}=\frac{\sin(z_k)}{-\sin(z_k)}=-1$. Then $\int_C f(z)dz=2\pi i\cdot\operatorname{Res}_{z=z_k}f(z)=-2\pi i$. But the answer says it is supposed to be $-4\pi i$, so what am I doing wrong?
Best Answer
You just forgot the other singulairty : $z=-\frac {\pi }2$. The residue at this point is also $-1$ so the integral is $(-1-1)(2\pi i)=-4\pi i$.