I need to find the residue of the following equation at $z = 0$.
$$ \frac{\cot(z)\cot(hz)}{z^3}$$
My attempt is as follows:
The residue will be the coefficient of $1/z$ in the Laurent Series expansion. Expansion of $$\cot(z) = 1/z - z/3 - z^3/45$$ and $$\cot(hz) = 1/z + z/3 - z^3/45 $$
sing these two we get coefficient of $1/z$ as $-2/45$.
Is this approach correct?
I assume that you actually mean $\coth z$, not $\cot(hz)$ (which is supported by your series expansions).
You need to multiply the two series:
$$ \Big(\frac1z - \frac13 z - \frac{1}{45}z^3 + O(z^5)\Big) \cdot \Big(\frac1z + \frac13 z - \frac{1}{45}z^3 + O(z^5)\Big) = \frac1{z^2} - \frac{7}{45}z^2 + O(z^4) $$
which shows that the residue of $\dfrac{\cot z \coth z}{z^3}$ at $z=0$ is equal to $-\dfrac{7}{45}$.
