<s>[INST] What is the residue of $z^2 \cos(\frac{1}{z})$ 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.
 [/INST] 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 $0$.
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