I need some help finding the Laurent expansion and residue of
$$\dfrac{\exp \left(\frac1z \right)}{(1-z)}$$
So far I've done
$$\sum_{j=0}^\infty \frac{z^{-j}}{j!} \sum_{k=0}^\infty z^k = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}$$
but don't know where to go from here. And is it also possible to use Cauchy product when one of the powers is $<0$ and the other is $>0$?
Best Answer
As for the residue: going "naive" may be a good idea$$\frac{1}{1-z}\,e^{\frac{1}{z}}=\left(1+z+z^2+z^3+...+z^n+...\right)\left(1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+...+\frac{1}{n!z^n}+...\right)$$
Well, it seems fairly easy to see what products are going to give us the coefficient of $\,z^{-1}\,$:
(first term left) times (second term right), (second left) times (third right),...,(n-th left) times ((n+1)-th right),..., so:$$\frac{1}{z}+\frac{1}{2z}+\frac{1}{6z}+...+\frac{1}{n!z}+...=\frac{1}{z}\sum_{n=1}^\infty\frac{1}{n!}=\frac{1}{z}(e-1)$$