Residue of $\cot z $ at $z= 0$ I need to calculate the Residue of $\cot z$ at the pole $0$ . Now, I think that the order of the pole is $1$. But I do not know how to show it. So, assuming the order of pole to be $1$, I calculated the residue as follows - $$ \lim_{z \rightarrow 0} (z - 0) \cot z  = \lim_{z \rightarrow 0} \frac{\cos z}{\frac{\sin z}{z}} = \frac{\lim_{z \rightarrow 0} \cos z}{\lim_{z \rightarrow 0} \frac{\sin z}{z}} = 1$$  Is my answer right? Again back to the main question, how do I determine or show the order of a pole(in general) ?
 A: Your answer is correct, and here is another way to try, in particular when taking derivatives or limits can be a little troublesome, and using the fact that we're only interested in low powers of $\;z\;$ in power or Laurent series since we want to find out something at $\;z=0\;$:
$$\frac{\cos z}{\sin z}=\frac{1-\frac{z^2}2+\ldots}{z-\frac{z^3}6+\ldots}=\frac{1-\frac{z^2}2+\ldots}{z\left(1-\frac{z^2}6+\ldots\right)}=\frac1z\left(1-\frac{z^2}2+\ldots\right)\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)=$$
You can see the above two parentheses are power series and thus analytic, so in order to find out what the coefficient of $\;z^{-1}\;$ is we just do
$$=\frac1z\left(1-\frac{z^3}3+\ldots\right)=\frac1z+\ldots$$
and thus the residue is certainly $\;1\;$ .
A: Yes your answer is right. 
In general, assume to have two holomorphic functions $f$, $g$ with a zero at $z_0$. Assume $z_0$ is a zero of multiplicity $p$ of $f$ and multiplicity $q$ of $g$, then $z_0$ is a pole of order $q-p$ of $f/g$. If $q-p \leq 0$, the singularity is removable.
Here $0$ is a zero of multiplicity $1$ of $\sin(z)$ (since it doesn't cancel the derivative) and is a zero of multiplicity $0$ of $\cos(z)$ (since it doesn't cancel the cosine itself). Thus, it is a pole of order $1-0$ of $\frac{\cos(z)}{\sin(z)}.$
