Coefficients of Laurent series

Question

I'm trying to find the coefficients $c_{-2}$, $c_{-1}$, $c_{0}$, $c_{1}$ and $c_{2}$ of the Laurent series around $z=0$ for the function $f(z)=\frac{e^z}{z(1-z)}$ in the region $1<|z|<\infty$. I've arrived at $$f(z)=-\sum_{n,m=0}^{\infty}\frac{1}{n!}z^{n-m-2}$$ but I don't know how to continue from where. Any help would be greatly appreciated.

Answer 1

First,

$$\frac1{z(1-z)}=\frac1z+\frac1{1-z}\implies \frac{e^z}{z(1-z)}=e^z\left(\frac1z+\frac1{1-z}\right)$$

so for $\;|z|>1\;$ we'll get

$$\sum_{n=0}^\infty\frac{z^n}{n!}\left(\frac1z-\frac1z\frac1{1-\frac1z}\right)=\frac1z\sum_{n=0}^\infty\frac{z^n}{n!}\left(1-\sum_{k=0}^\infty z^{-k}\right)=$$

$$\sum_{n=0}^\infty\frac{z^{n-1}}{n!}-\sum_{n=0}^\infty\frac{z^{n-1}}{n!}\sum_{k=0}^\infty z^{-k}=$$

$$=\frac1z+1+\frac z2+\frac{z^2}6+\ldots-\left(\frac1z+1+\frac z2+\frac{z^2}6+\ldots\right)\left(1+\frac1z+\frac1{z^2}+\frac1{z^3}+\ldots\right)=$$

$$=\sum_{n=0}^\infty\frac{z^{n-1}}{n!}-\sum_{n,k=0}^\infty\frac{z^{n-k-1}}{n!}$$

Now, $\;c_{-2}\;$ comes only from the second summand, and it happens when $\;n-k-1=-2\implies n-k=-1\;$, so if we denote the pairs of non-negative integers that fulfill this as $\;n,k\;$ , we'll get $\;(0,1), (1,2), (2,3),...,(h, h+1),\ldots\;$, and thus we get (observe that only $\;n\;$ has an actual value here...)

$$c_{-2}=-\frac1{0!}+\frac1{1!}+\frac1{2!}+\ldots+\frac1{n!}+\ldots=\sum_{n=0}^\infty\frac1{n!}=-e$$

For $\;c_{-1}\;$ : the first summand contributes with $\;\frac1{0!}=1\;$ for $\;n=0\;$ , and in the second sumand we have that $\;z^{n-k-1}=z^{-1}\iff z=n\;$, so there:

$$\frac1{0!}+\frac1{1!}+\ldots\frac1{n!}+\ldots\implies c_{-1}=1-\sum_{n=0}^\infty\frac1{n!}=1-e$$

Well, now you try the other coefficients you need...