Compute the residue of $ 1/(z^2 \sin z) $ at $z=0$

Question

I'm trying to compute the following:

$$ \operatorname*{Res}_{z=0} \frac{1}{z^2 \sin z} $$

I know I can either find an analytic function and use the equation $$ \operatorname*{Res}_{z=z_0} \frac{p(z)}{q(z)}=\frac{p(z_0)}{q(z_0)} $$ so long as $p(z_0)\ne0, \, q(z_0)=0, \, q'(z_0)\ne0 $, or figure out some new expansion.
My first attempt at this was to choose $\csc(z)$ as my analytic function, $p(z)$, and $z^2$ as $q(z)$. But $q'(0)=0$, which fails to satisfy the requirements of the equation.

I could really just use a hint to get started, I'm not sure how else to approach this without doing an expansion.

Any help would be appreciated, thank you.

Answer 1

Elaborating on the comment by Daniel Fischer:

$$\begin{align} f(z) &= \frac{1}{z^2\sin z} = \frac{1}{z^3(1-\frac{z^2}{6}+O(z^4))}=\frac{1}{z^3}(1-\frac{z^2}{6}+O(z^4))^{-1}\\ &=\frac{1}{z^3}(1+\frac{z^2}{6}+O(z^4)) = \frac{1}{z^3}+\frac{1}{6z}+O(z). \end{align}$$

Where I have used the expansion $(1+z)^\alpha = 1 +\alpha z +\frac{\alpha(\alpha-1)}{2}z^2 +\dots$ for $\alpha=-1$. The coefficient of the $1/z$ term is the residue, which in this case is $1/6$.

Because $z^3f(z)$ is for this function analytic at $z=0$, the function has a pole of order 3. You could also use the general formula for poles of order $m$, with $m=3$ to get the residue at $z=0$.

$$ \mathrm{Residue} = \lim_{z\to0}\frac{1}{2}\frac{d^2}{dz^2}z^3f(z),$$

which after a longer calculation will also give you 1/6 (easier and shorter with the expansion suggested by Daniel Fisher).

Answer 2

Here is an alternative derivation without using a series expansion.

You only need that $\frac {1-\cos z} {z^2} = 1/2$ and that $\lim_{z \to 0}\frac z {\sin z} =1$. These limits are known, as they are equivalent to to $\frac 1 2\frac {d} {dz^2} \cos(z)|_{z=0}$ and $\frac d {dz} \sin(z)|_{z=0}$.

Specifically, the latter implies that $z^2 \sin z$ has a zero of order 3 at $0$. Thus, (by definition of the residue) $$\operatorname{Res}_{z=0}(\frac 1 {z^2 \sin(z)})= \frac 1 {(3-1)!} \lim_{z\to 0} \frac {d^{3-1}}{dz^{3-1}}\frac {z^3}{z^2 \sin z}$$ Therfore, $$\operatorname{Res}_{z=0}(\frac 1 {z^2 \sin(z)})= \frac 1 2 \lim_{z\to 0} \frac {d^{2}}{dz^{2}}\frac {z}{\sin z} $$

We compute the second derivative of $f(z) = \frac z{\sin z}$ directly from definition. This gives $$\frac 1 2 \lim_{z\to 0} \lim_{h\to 0} \frac{f(z+h) - 2f(z) +f(z-h)}{h^2} ~.$$ Exchanging limits and using continuity + symmetry of $f$ yields $$\frac 1 2 \lim_{h\to 0} \frac{2 f(h) - 2f(0)}{h^2} = \lim_{h\to 0} \frac{ \frac{h}{\sin h} - 1}{h^2} = \lim_{h\to 0} \frac{ h - \sin h}{h^2\sin h} = \lim_{h\to 0} \frac{ h - \sin h}{h^2\sin h}~.$$

This limit can finally be computed with l'Hospital rule. $$\lim_{h\to 0} \frac{ h - \sin h}{h^2\sin h} = \lim_{h\to 0} \frac{ 1 - \cos h}{2 h\sin h + h^2 \cos h} \\ = \lim_{h\to 0} \frac{ 1 - \cos h}{h^2} \lim_{h\to 0}\frac {1}{2 \frac {\sin h}{h} + \cos h} = \frac{1}{2} \frac{1}{3} = \frac 1 6$$