Laurent Expansion of a given function

Question

$$f(z)=\frac{z^2-2}{z^3}\cos\left(\frac{1}{z+1}\right)$$ I suppose $f(z)$ has an essential singularity at $z=-1$. I found the Laurent expansion to be $$\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^{n}[2.C(m+2,m)-1]}{(2n)!}(z+1)^{m-2n}$$ Is this right? If wrong then what will be the Laurent expansion of $f(z)=\frac{z^2-2}{z^3}\cos\left(\frac{1}{z+1}\right)$ ?

Answer 1

Yes, $f$ has an essential singularity at $-1$. If you compute the Taylor series of $\frac{z^2-2}{z^3}$ centered at $-1$, which turns out to be $\sum_{n=0}^\infty(n^2+3n+1)(z+1)^n$. And you have$$\cos\left(\frac1{z+1}\right)=1-\frac1{2!(z+1)^2}+\frac1{4!(z+1)^4}-\cdots$$Therefore\begin{align}f(z)&=\frac{z^2-2}{z^3}\cos\left(\frac1{z+1}\right)\\&=\left(1+5(z+1)+11(z+1)^2+\cdots\right)\left(1-\frac1{2!(z+1)^2}+\frac1{4!(z+1)^4}-\cdots\right)\end{align}and you can use this to comput the Laurent series of $f$ centered at $-1$, although I don't think that there is a nice closed expression for it. For instance, the constant term will be$$1-\frac{11}{2!}+\frac{29}{4!}-\cdots$$whereas the coefiicient of $(z+1)^{-1}$ (that is, the residue of $f$ at $-1$) will be$$-\frac5{2!}+\frac{19}{4!}-\frac{41}{6!}+\cdots$$