Laurent Series $e^{\frac1{1-z}}$, $|z|>1$.

Question

I tried to find Laurent expansion for:

$e^{\frac1{1-z}}$, $|z|>1$.

I tried next: $\frac1{1-z}=-\sum_{n=1}^{\infty}\frac1{z^n}$,

then using $e^{\frac1{1-z}}=1+\frac1{1-z}+\frac{1}{{2!(1-z)}^2}+\frac{1}{{3!(1-z)}^3}+...$

Then I have $e^{\frac1{1-z}}$ = $1$ + $(-\frac1{z}-\frac1{z^2}-\frac1{z^3}-...) + (\frac1{2!})(\frac1{z^2}+\frac1{z^4}+…)+...$ = $1 - \frac1z-\frac1{2z^2}-...$.

The result is wrong, since in the book the answer is: $1 - \frac1z+ \frac1{2z^2}-\frac1{6z^3}+\frac1{24z^4}-\frac{19}{120z^5}...$

What I did wrong?

Answer 1

With $z=\dfrac{1}{w}$ we have $|w|<1$ then by $e^z$ expansion \begin{align} e^{\frac{1}{1-z}} &= e^{\frac{-w}{1-w}} \\ &= \sum_{n\geq0}\dfrac{1}{n!}\left(\frac{-w}{1-w}\right)^n \\ &= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1-w\right)^{-n} \\ &= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1+nw+\dfrac{n(n+1)}{2}w^2+\cdots\right) \\ &= \sum_{n\geq0}\dfrac{(-1)^n}{n!}\left(w^n+nw^{n+1}+\dfrac{n(n+1)}{2}w^{n+2}+\cdots\right) \end{align} after calculation some terms we let $w=\dfrac1z$.

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Edit: One may find \begin{align} &1 \\ &-(w+w^2+w^3+w^4+w^5+\cdots) \\ &+\frac12(w^2+2w^3+3w^4+4w^5+\cdots) \\ &-\frac16(w^3+3w^4+6w^5+\cdots) \\ &+\cdots\\ &=1-w-\frac{w^2}{2}-\frac{w^3}{6}+\frac{w^4}{24}+\frac{19 w^5}{120}+\cdots\\ &=\color{blue}{1-\dfrac1z-\frac{1}{2z^2}-\frac{1}{6z^3}+\frac{1}{24z^4}+\frac{19}{120z^5}+\cdots} \end{align}

Answer 2

Note that$$\left(-\frac1z-\frac1{z^2}-\frac1{z^3}-\frac1{z^4}-\cdots\right)^2=\frac1{z^2}+\frac2{z^3}+\frac3{z^4}+\cdots,$$that$$\left(-\frac1z-\frac1{z^2}-\frac1{z^3}-\frac1{z^4}-\cdots\right)^3=-\frac1{z^3}-\frac3{z^4}-\frac6{z^5}-\cdots,$$and so on. This explains why you got the wrong result.