Determining the residue of a function by computing its laurent expansion

Question

I need to find the residue of $$f(z) = \frac{e^z\sin(z)}{z(1-\cos(z))}$$ at $z = 0$ via its Laurent series expansion. First of all, I tried to expand all the functions via Taylor series at $z = 0$. I obtained, after some easy computations, the following expression: $$\frac{2}{z^2}\cdot \frac{1+z+\frac{z^2}{3}+\dots}{1-\frac{z^2}{12}+\dots }$$ Now, my idea was to somehow simplify the denominator, by multiplicating and dividing by some factor, and get an expression of this form: $\frac{2}{z^2} (1+z+\frac{z^2}{3}+\dots)(?\dots ?)$. I failed and couldn't find a way to put the above expression in the usual form of a Laurent series, namely: $\sum_{n=-\infty}^{+\infty}c_nz^n$ since I only need the $c_{-1}$ coefficient in order to determine the residue If someone could give me a hand, I'd gladly appreciate it.

Answer 1

The first terms for $z\sin z$ are $$e^z\sin z=z+z^2+\frac{z^3}{3}-\frac{z^5}{30}-\frac{z^6}{90}+O(z^7)+\cdots,$$ and the first terms for $$1-\cos z=-\frac{z^2}{2}+\frac{z^4}{4!}+\frac{z^6}{6!}+\cdots.$$ Since the pole at $z=0$ of $\frac{1}{1-\cos z}$ has order $2$ we know that the laurent series is on the form $\sum_{n=-2}^\infty a_n z^n,$ with the odd terms equal to zero, so $$\left(-\frac{z^2}{2}+\frac{z^4}{4!}+\frac{z^6}{6!}+\cdots\right)\left(a_{-2}z^{-2}+a_0+a_2z^2+\cdots\right)=1,$$ it's easy to compute the first terms, where you obtain that $$\frac{1}{1-\cos z}=\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120}+\cdots.$$ Now we have that $$\frac{1}{z}\frac{e^z\sin z}{1-\cos z}=\left(1+z+\frac{z^2}{3}+\cdots\right)\left(\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120}+\cdots\right)=\frac{2}{z^2}+\frac{2}{z}+\frac{5}{6}+\frac{z}{6}+\cdots,$$ so the residue is $2$.

Answer 2

First we simplify it

$$f(z) = \frac{e^z\sin(z)}{z(1-\cos(z))}=\frac{e^z\cdot2\sin(\frac z2)\cos(\frac z2)}{z\cdot2\sin^2(\frac z2)}=\frac{e^z\cdot\cos(\frac z2)}{z\cdot\sin(\frac z2)}$$

We can see, $z=0$ is a second order pole from above result. Then by residue formula

$$\lim_{z\to0}\frac{d}{dz}z^2f(z)=2$$

Or, if you want to do the Laurent series and find the coefficient of $z^{-1}$ term, you can do the following,

$$\frac{e^z\cdot\cos(\frac z2)}{z\cdot\sin(\frac z2)}\sim\frac{1}{z\cdot \frac z2}\cdot \left(1+z+O(z^2)\right)\cdot\left(1+O(z^2)\right)$$

Since we only need the $z^{-1}$ term, hence

$$\frac{1}{z\cdot \frac z2}\cdot z=2\cdot\frac1z$$

the coefficient is $2$, which agrees with above result by applying residue formula.