Residue at infinity of $f(z)=z^3\cos\big(\frac{1}{z-2}\big)$

Question

I have trouble with the residue of: $f(z)=z^3\cos\left(\frac{1}{z-2}\right)$ at $z = \infty$. I tried to solve it at $z=0$ but it turns out that I was wrong while $z=0$ is not a pole. I must solve it at $z=2$ but I'm stuck. Any suggestion will be much appreciated.

Answer 1

The residue at infinity of a function $f$ holomorphic in a punctured neighbourhood of $\infty$ is the residue in $0$ of the function

$$g(z) = -\frac{1}{z^2}f\left(\frac1z\right).$$

For $f(z) = z^3\cos \frac{1}{z-2}$, that becomes the residue in $0$ of

$$-\frac{1}{z^5}\cos \frac{z}{1-2z}.$$

Expanding $\frac{z}{1-2z}$ into a geometric series and inserting the result into the Taylor series of $\cos$ yields

$$\begin{align} \frac{z}{1-2z} &= z + 2z^2 + 4z^3 + O(z^4)\\ \cos (z + 2z^2 + 4z^3 + O(z^4)) &= 1 - \frac12(z+2z^2+4z^3)^2 + \frac{1}{24}z^4 + O(z^5)\\ &= 1 - \frac12 z^2 - 2z^3 - \frac{143}{24}z^4 + O(z^5), \end{align}$$

so the residue is $\frac{143}{24}$.