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Give an explicit formula for the coefficients of the Laurent series on $A:=\{z: |z|>1\}$ for the function $g(z)=\frac{e^z}{z-1}$.

I know how to go about finding the Laurent series, using the geometric series for $\frac{1}{1-z}$. But how do I obtain an explicit formula for ALL the coefficients?.

I have seen some earlier posts,but I am not entirely clear on their work. Can we use the coefficient formula in the Laurent series expansion and then use Cauchy's formula?.

Any help is appreciated.

2 Answers 2

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You can at first determine the Laurent series for $ e^{z} = \sum_{m=0}^\infty \frac{z^m}{m!}$ and $ \frac1{z-1} = \sum_{n=0}^\infty z^{-(n+1)} , \qquad |z|>1$ independently.

To multiply two Laurent series, we can use this formula or simply calculate $\begin{align}\frac{e^{z}}{z-1}= &\left(\sum_{m=0}^\infty \frac{z^m}{m!} \right) \left(\sum_{n=0}^\infty z^{-(n+1)}\right)\\ &=\sum_{m,n\in\mathbb{Z}}[0\leq m][0\leq n] \frac{z^{m-n-1}}{m!}\\ &=\sum_{k,m} [0\leq m][0\leq m-k-1] \frac{z^{k}}{m!}\\ &= \sum_{k=-\infty}^{-1} z^k \underbrace{\sum_{m=0}^\infty \frac1{m!}}_{e} + \sum_{k=0}^\infty z^k \sum_{m=1+k}^\infty \frac1{m!}. \end{align}$ with $k=m-n-1$ and where I used Iverson's bracket.

Edit: As Robert Israel pointed out, we can still express the last sum using the incomplete $\gamma$-function $\gamma(n,x) = \int_0^x t^{n-1} e^{-t} dt = x^n (n-1)! e^{-x} \sum_{k=0}^\infty \frac{x^k}{(n+k)!} .$ Setting $x=1$, we have with $n\geq 1$ $\gamma(n,1) = (n-1)! \sum_{i=n}^\infty \frac{1}{i!}. $ Thus, $\frac{e^z}{z-1} = e \sum_{k=-\infty}^{-1} z^k+ \sum_{k=0}^\infty\frac{\gamma(k+1,1)}{k!} z^k.$

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    @user54755: Laurent expansion are always in rings about a centre. If you expand in powers of $z-1$ you get a laurent expansion in a ring centered at $1$. But the question obviously asks for a Laurent expansion in a ring centered at 0.2013-01-01
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First observe $\frac{1}{1-z}=-\frac1z\frac{1}{1-\frac1z}=\sum_{-\infty}^{n=-1}-z^n$ and$e^z=\sum_{k=0}^{\infty}\frac{z^k}{n!}$ The general term of the first series is $a_n=0$, $n\ge 0$ and $a_n=-1$, $n<0$ while for the second $b_n=\frac1{n!}$ for $n\ge 0$ and $b_n=0$ for $n<0$. $\sum_{n=-\infty}^{\infty}a_n\sum_{n=-\infty}^{\infty}b_n=\sum_{n=-\infty}^{\infty}c_n$ where $c_n=\sum_{k=-\infty}^{\infty}a_kb_{n-k}$ Can you continue?

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    @user54755 By expanding at $0$ I mean at A=\left\{\left|z\right|>1\right\} which is centered at $0$ and the expansion is valid as in my answer.2013-01-01