What is the residue of $z^2 \cos(\frac{1}{z})$

Question

Can I find it using Laurent series expansion? I tried using the Taylor series expansion of $\cos(1/z)$ but could not reach any conclusion.

Answer 1

The residue is the coefficient of the term $\frac{1}{z}$ in the Laurent Series.

The Maclaurin Series of $\cos z$ is

$$\cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots $$

It follows that

$$\cos\left(\frac{1}{z}\right) = 1 - \frac{1}{2!}\frac{1}{z^2} + \frac{1}{4!}\frac{1}{z^4} - \frac{1}{6!}\frac{1}{z^6}+\cdots $$

Finally, multiplying both sides by $z^2$, we get

$$z^2\cos\left(\frac{1}{z}\right) = z^2 - \frac{1}{2!} + \frac{1}{4!}\frac{1}{z^2}- \frac{1}{6!}\frac{1}{z^4}+\cdots $$

The coefficient of $\frac{1}{z}$ is $0$, so the residue of $z^2\cos\left(\frac{1}{z}\right)$ is $0$.

Answer 2

I assume that you find the expansion around $z=0$. Just find the Laurent series of $\cos(z)$ and substitute by $z^{-1}$, then multiply both sides by $z^2$ to get the desired result. Then use residue theorem to find the answer.