Let $f(z) =e^{\left(\frac{1}{(1+z)^2}\right)} $. How could I calculate the Laurent series of this function around 0 and -1?
I know how to calculate the Laurent series of $e^{(1/z)}$ and of $\frac{1}{(1+z)^2}$ however here I do no not know how to proceed.
Edit: Around - 1: would it be like this?
$e^{\left(\frac{1}{(1+z)^2}\right)} = \sum_{n=0}^{\infty} \frac{{(\frac{1}{(1+z)^2})}^{n}}{n!} $
If $f(z)$ and $g(z)$ are power zeries, you cannot work out the composition $g(f(z))$ by substituting one series into the other unless the constant term in $f(z)$ is 0. If it is 0, you need first to apply an addition formula to the series $g(z)$ to knock off the 0 term. Thus if we write $$f(z)=a_0+a_1z+a_z++a_2z^3+...$$ and $$h(z)=a_1z+a_z++a_2z^3+...$$ we have$$\exp(a_0+a_1z+a_z++a_2z^3+...)=e^{a_0}\exp(h(z))$$ $$=e^{a_0}(1+\sum_{n=1}^{\infty}\text{coefficient of $z^n$ in} \sum_{m=1}^n\frac{1}{m!}(h(z)^m) )$$. Around $z=0$ we have $$\frac{1}{(1+z)^2}=(1-z+z^2-z^3+...)^2$$ $$=1-2z+3z^2-4z^3+...$$. Now we can apply the formula above with $a_0=1$ and $h(z)=-2z+3z^2-4z^3+...$ $$\text{Around $z=-1$ we have the otherkind of Laurent series,}$$ in which we have at most finitely many terms raised to a non-negative power, giving us $$\exp(\frac{1}{(1+z)^2}=1+\frac{1}{(z+1)^2}+\frac{1}{2!}(\frac{1}{(z+1)^4})+\frac{1}{3!}(\frac{1}{(z+1)^6})+...$$