Difficult residue $Res(z+2)^2{cos{z\over{1+z}}}$ at $z=-1$

Question

$Res( (z+2)^2{cos{z\over{1+z}}})$ at $z=-1$ Usually calculating residues is easier. In this case the point is essential singularity,so I tried to expand cos into series, and expand $ z\over{1+z} $ and combine the series to find $a_{-1}$, but it gets difficult to calculate.

Answer 1

Let $\zeta=z+1$. Then the Laurent expansion of the function is

$$\begin{align}(1+\zeta)^2 \cos{\left ( 1-\frac1{\zeta} \right )} &= (1+2 \zeta+\zeta^2) \left (\cos{(1)} \, \cos{\left ( \frac1{\zeta} \right )} + \sin{(1)} \, \sin{\left ( \frac1{\zeta} \right )} \right )\\ &= (1+2 \zeta+\zeta^2) \left [\cos{(1)} \sum_{n=0}^{\infty} \frac{(-1)^n \zeta^{-2 n}}{(2 n)!} + \sin{(1)} \sum_{n=0}^{\infty} \frac{(-1)^n \zeta^{-2 n-1}}{(2 n+1)!} \right ] \end{align}$$

The residue at $\zeta=0$ is the coefficient of $\zeta^{-1}$. We can then read this right off from the sum as

$$2 \cos{(1)} \left (-\frac12 \right ) + \sin{(1)} \left (1 - \frac16 \right ) = \frac56 \sin{(1)} - \cos{(1)}$$