I need to calculate residue of $\cot^2(z)$ at all poles. But I don't know how to solve it.
I am trying to solve in this way:
$$\lim_{z\to a} \frac{d}{dz}\left(\cot^2(z)(z-a)^2\right)$$ where $a \in \{k\pi: k \in Z\}$
But I don't know how to solve this limit of derivation.
Can anybody help me, please?
Best Answer
$$\cot^2(z) = \frac{\cos^2(z)}{\sin^2(z)} \ \ \text{ so we need only worry about the zeros of } \sin^2(z)$$
These zeros occur when $z_k = k \pi$ for $k \in \mathbb{Z}$, as you've already identified.
Moreover, they are 2nd order zeros, as $(\sin^2(z))' = 2 \sin(z) \cos(z)$, which also has zeros at $z_k$, whereas $(2 \sin(z) \cos(z))' = 2 \cos^2(z) - 2\sin^2(z)$, which is nonzero at $z_k$.
Hence, $\cot^2(z)$ has $2^{nd}$ order poles when $z_k = k \pi, k \in \mathbb{Z}$. Therefore,
$$\text{Res}_{z = z_k} \cot^2(z) = \lim_{z \to z_k} \frac{d}{dz}\frac{\cos^2(z)(z - z_k)^2}{\sin^2(z)}$$
Calculate this derivative, then take the limit by applying L'Hopital's rule
In general, if $f(z)$ has an $n^{th}$ order pole at $z = z_k$, then
$$\text{Res}_{z = z_k} f(z) = \frac1{(n-1)!} \lim_{z \to z_k} \frac{d^{n-1}}{dz^{n-1}} \left[(z-z_k)^n\, f(z)\right]$$
Which can be seen directly from the Laurent expansion of $f$