Calculate the residue of this function 
Find the residue at $z=0$ of the function $f(z)=\frac{\cot z}{z^4}$

I know that $z_0=0$ is a pole of order $k=5$, and $$Res(f;z_0)=\frac{\phi(z_0)^{(k-1)}}{(k-1)!}$$
but I cannot get the right answer that is $-\frac{1}{45}$

Laurent series approach It is easy to see that the Laurent series of $\cot(z)$ around $z=0$ is 
$$
\cot(z)=\frac 1z - \frac z3 - \frac{z^3}{45} - \cdots 
$$
Thus 
$$
\frac{\cot(z)}{z^4}=\frac{1}{z^5} - \frac{1}{3z^3} - \frac{1}{45z} - \cdots 
$$
Hence, as pointed out in the comments the order of the pole is $5$ and indeed $Res(\cot; 0)=a_{-1}=-1/45$
Without using Laurent Series It is more complicated since you need to compute 4 derivates. If you insist you will need to do as follows: because the pole is of order $5$, then 
\begin{align}
Res(\cot; 0)=a_{-1}=\frac{1}{4!}\lim_{z \to 0}\frac{d^4}{dz}\left( z^5 \frac{\cot(z)}{z^4}\right) & =\frac{1}{24}\lim_{z \to 0}\frac{d^4}{dz}\left( z\cot(z)\right)\\ 
 & = \left(\frac{1}{24}\right)\left(\frac{-8}{15}\right)\\
& = -\frac{1}{45}
\end{align}
However it is a little bit messy (click here too see why $\lim d^4(z\cot(z))=-8/15$ ). For this case, I of course prefer the Laurent series option.
