Evaluate the integral $\int_{C}\frac{\cos(z)}{\sin(z)}dz$, where $C$ is a circle centered at the origin and there is no zero of $\sin(z)$ inside of $C$ other than the point $(0,0)$.
I think for this problem I need to apply the residue theorem and clearly the origin is a pole. But how do I determine the order of this pole? Or there is another way to solve this problem without the residue theorem? Thanks.
Note that $\lim_{z\to 0}\frac{z}{\sin(z)}=1$. Hence, $z=0$ is a pole of order $1$.
So, we have
$$\begin{align} \oint_{|z|=r<\pi}\frac{\cos(z)}{\sin(z)}\,dz&=2\pi i \text{Res}\left(\frac{\cos(z)}{\sin(z)}, z=0\right)\\\\ &=2\pi i \lim_{z\to 0}\frac{z\cos(z)}{\sin(z)}\\\\ &=2\pi i \end{align}$$
Another way to see that $\cot(z)$ has a pole of order $1$ is to write
$$\begin{align} \cot(z)&=\frac{\cos(z)}{\sin(z)}\\\\ &=\frac{1-\frac12z^2+O(z^4)}{z-\frac16z^3+O(z^5)}\\\\ &=\frac{1-\frac12z^2+O(z^4)}{z\left(1-\frac16 z^2+O(z^4)\right))}\\\\ &=\frac1z\left(1-\frac13z^2+O(z^4)\right)\\\\ &=\frac1z-\frac13z+O(z^3) \end{align}$$
whereupon we see from the first term of the Laurent series that $z=0$ is a pole of order $1$ with residue $1$.