How can I classify the singularity in $z = 0$ and determine the respective residue in $z = 0$ for the following function ?
f(z) = $ cos(1/z)(z+1)^2$
Do I have to use Taylor expansion of $cos(1/z)$ and $(z+1)^2$ ? If yes, then what? If no, how should I begin?
Thanks!
$$\cos(1/z)(1+z^2)=(z^2+2z+1)\cos(1/z)=(z^2+2z+1)\sum_{n=0}^{\infty}(-1)^{n}\frac{(1/z)^{2n}}{(2n)!}=\\=(z^2+2z+1)\sum_{n=0}^{\infty}(-1)^{n}\frac{1}{z^{2n}(2n)!}=\\=\sum_{n=0}^{\infty}(-1)^{n}\frac{1}{z^{2n-2}(2n)!}+\sum_{n=0}^{\infty}(-1)^{n}\frac{2}{z^{2n-1}(2n)!}+\sum_{n=0}^{\infty}(-1)^{n}\frac{1}{z^{2n}(2n)!}$$