Residue of $f(z)=\frac{\cot(z)\coth(z)}{z^3}$ How can I find the residue of the following function at the point $z=0$

$$f(z)=\frac{\cot(z)\coth(z)}{z^3}$$

 A: Keep in mind that each of $\cot{z}$ and $\text{coth}{z}$ has a simple pole at $z=0$.  Thus $z=0$ is a pole of order $5$, and the residue is equal to
$$\frac{1}{4!} \left [ \frac{d^4}{dz^4} \left ( z^2 \cot{(z)} \, \text{coth}(z) \right) \right ]_{z=0}$$
ADDENDUM
The above expression, while correct, is not really practical without access to a computer algebra software package.  (I'm sure in the olden days, someone had to work out something like this by hand.  But...why?)  Better to work directly with the Taylor expansions of $z \cot{z}$, etc. about $z=0$ directly as follows:
$$z \cot{z} = 1-\frac{z^2}{3} - \frac{z^4}{45}+ O\left (z^6\right)$$
$$z \coth{z} = 1+\frac{z^2}{3}  - \frac{z^4}{45}+ O\left (z^6\right)$$
So we want the coefficient of $1/z$ in the product of the above divided by $z^5$.  This coefficient is straightforward as the sum of three terms:
$$-\frac{1}{45} - \frac{1}{45} - \frac13 \cdot \frac13 = -\frac{7}{45}$$
This of course agrees with the result of taking the 4th derivative above, but is much easier to see.
