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I am to obtain the first 3 terms of the Laurent series about $z = 0$ for:

$$f(z)=1/(e^z-1-z)$$

I know that there is 1 singularity at $z=0$. My denominator is not in a polynomial form so I can convert it into such by taking a Taylor series of the function around z=0 which yields the denominator as:

$$(e^z-1-z) = x^2/2 + x^3/6 + x^4/24 + ...$$

This is the part where I get stuck. I'm supposed to bring that back in and perhaps factor something out of the denominator so the left side is analytic at z=0 and the singularities are shifted to the term on the right side. Any tips on what my next step should be? Many thanks!

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Since $$ e^{z}-1-z=z^2\sum_{n=0}^\infty\frac{z^n}{(n+2)!}\tag1 $$ let $$ \frac{z^2}{e^{z}-1-z}=\sum_{k=0}^\infty a_kz^k\tag2 $$ Then using the Cauchy Product Formula $$ \begin{align} 1 &=\sum_{n=0}^\infty\frac{z^n}{(n+2)!}\sum_{k=0}^\infty a_kz^k\\ &=\sum_{n=0}^\infty\sum_{k=0}^n\frac{a_{n-k}}{(k+2)!}z^n\tag3 \end{align} $$ That is, equating the coefficients of the powers of $z$, $$ \sum_{k=0}^n\frac{a_{n-k}}{(k+2)!}=[n=0]\tag4 $$ where $[\cdots]$ are Iverson Brackets.

Plugging $n=0$ into $(4)$, we get $a_0=2$.

Solving for $a_n$, for $n\ge1$, in terms of previous terms, we get $$ a_n=-2\sum_{k=1}^n\frac{a_{n-k}}{(k+2)!}\tag5 $$ So we can compute all the $a_n$ iteratively using $(5)$.

Then we get the Laurent series: $$ \begin{align} \frac1{e^z-1-z} &=\sum_{k=0}^\infty a_kz^{k-2}\\ &=\frac2{z^2}-\frac2{3z}+\frac1{18}+\frac{z}{270}-\frac{z^2}{3240}-\frac{z^3}{13608}-\cdots\tag6 \end{align} $$

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Since the Taylor series of $e^z-1-z$ begins with the term of degree $2$, the Laurent series that you're after is something like$$\frac{a_{-2}}{z^2}+\frac{a_{-1}}z+a_0+a_1z+\cdots$$So, you have\begin{align}1&=\left(\frac{z^2}2+\frac{z^3}6+\frac{z^4}{24}+\cdots\right)\left(\frac{a_{-2}}{z^2}+\frac{a_{-1}}z+a_0+a_1z+\cdots\right)\\&=\left(\frac12+\frac z6+\frac{z^2}{24}+\cdots\right)\left(a_{-2}+a_{-1}z+a_0z^2+a_1z^3+\cdots\right)\\&=\frac{a_{-2}}2+\left(\frac{a_{-1}}2+\frac{a_{-2}}6\right)z+\left(\frac{a_0}2+\frac{a_{-1}}6+\frac{a_{-2}}{24}\right)z^2+\cdots\end{align}and so all you need is to solve the system$$\left\{\begin{array}{l}\frac{a_{-1}}2=1\\\frac{a_{-1}}2+\frac{a_{-2}}6=0\\\frac{a_0}2+\frac{a_{-1}}6+\frac{a_{-2}}{24}=0.\end{array}\right.$$

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Another approach:

Using the power series of $\;e^z\;$ , we get:

$$e^z-1-z=\sum_{n=0}^\infty\frac{z^n}{n!}-1-z=\frac{z^2}{2}+\frac{z^3}{6}+\ldots$$

Since we're interested in $\;z\;$ "close" to zero (meaning $\;|z|<1\;$) , we get:

$$\frac1{e^z-1-z}=\frac1{\frac{z^2}2\left(1+\frac z3+\frac{z^2}{12} +\ldots\right)}=\frac2{z^2}\left(1-\frac z3-\frac{z^2}{12}+\left(\frac z3+\frac{z^2}{12}\right)^2-\ldots\right)=$$

$$=\frac2{z^2}-\frac{2}{3z}-\frac16+\frac29-\ldots=\frac2{z^2}-\frac2{3z}+\frac1{18}+\ldots$$

How to know how many summands to take in the development of $\;\frac1{1+\frac z3+\frac{z^2}{12}+\ldots}\;$ ? As many as required, and since we need only the first three elements and zero is a double pole, we only need to evaluate the coefficients of $\;z^{-2},\,z^{-1}\,,\,z^0\;$ ...

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Borrowing from that answer I commented, we have \begin{align} \frac{1}{e^z - 1 - z} &= \frac{1}{(1 + z + z^2/2! + \dots) - 1 - z}\\ &= \frac{1}{z^2/2! + z^3/3! + z^4/4! + \dots}\\ &= \frac{2}{z^2} \cdot \frac{1}{1 + 2z/3! + 2z^2/4! + \dots} \end{align} Now letting $P(z) = 2z/3! + 2z^2/4! + \dots$, we have \begin{align} \frac{1}{e^z - 1 - z} &= \frac{2}{z^2} \cdot (1 - P(z) + P(z)^2 - P(z)^3 + \dots)\\ &= \frac{2}{z^2} \cdot (1 - 2z/3! + z^2(-2/4! + (2/3!)^2) + \dots)\\ &= 2z^{-2} - \frac{2z^{-1}}{3} + \frac{1}{18} - \dots \end{align} Getting higher order terms starts to get a bit complicated, so I'd suggest using Wolfram Alpha to fill in the rest: http://www.wolframalpha.com/input/?i=laurent+series+of+1%2F(e%5Ez+-+1+-+z)

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