Calculate $\int_{\gamma}\cot z dz$ where $\gamma=C(0,3)$

Question

I've just started doing complex integrals with residues, and I'm struggling a bit with finding the singularities.

The first integral is:

$$\int_{\gamma}\cot z dz$$ where $\gamma=C(0,3)$.

My solution so far is:

1. Find singularities and determine what kind of singularity

   $\int_{\gamma}\cot z dz = \int_{\gamma}\frac{\cos z}{\sin z} dz$ where $z=0$ and $z=\pi k$ is singularites.

   Taylor expansion gives me $$\int_{\gamma}\frac{z(\frac{1}{z}-\frac{z}{2!}+\frac{z^3}{4!}...)}{z(1-\frac{z^2}{2!}+\frac{z^4}{4!}...)}dz = \int_{\gamma}\frac{\frac{1}{z}-\frac{z}{2!}+\frac{z^3}{4!}... }{1-\frac{z^2}{2!}+\frac{z^4}{4!}... }dz$$ $\Rightarrow$ pole of order one at $z=0$

   I also know there is a singularity $\pi k$, but don't know how to compute that.

2. Residue won't be a problem once I find the singularities.

Would anyone like to help me to give a clear explanation on how to find the singularity at $\pi k$?

Answer 1

So $\cot z$ has ploes at $z=k\pi$, where $k\in\mathbb{Z}.$

So in $C(0,3)$ you will have only one pole at $z=0$. Therefore using the residue theorem:

$$\int_{\gamma}\cot z dz=\int_{\gamma}\frac{\cos z}{\sin z} dz=2\pi i\ \text{Res}(\cot z, 0)=2\pi i.$$