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Given the function $f(z)=\exp\left(\frac{z}{1-z}\right)$, I want to find the coefficients $a_0$, $a_{-1}$, and $a_{-2}$ of the Laurent expansion $f(z)=\sum_{n=-\infty}^{\infty}a_n(z+1)^n$ about $z=-1$, on the annulus $\{z\in\mathbb{C}:|z+1|>2\}$.


We know that the power series for $\exp(z)$ centered about $z=-1$ is $\sum_{n=0}^{\infty}\frac{(z+1)^n}{e\cdot n!}$, so \begin{align*} \exp\left(\frac{z}{1-z}\right) &= \sum_{n=0}^{\infty}\frac{1}{e\cdot n!}\left(\frac{z}{1-z}+1\right)^n\\ &= \sum_{n=0}^{\infty}\frac{1}{e\cdot n!}\left(\frac{1}{1-z}\right)^n\\ &= \sum_{n=-\infty}^0\frac{(-1)^n}{e\cdot (-n)!}(z-1)^n \end{align*}

But this gives us the Laurent series of f(z) centered at 1 which does not seem to be helpful.

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  • $\begingroup$ I would not call that an annulus. $\endgroup$ Apr 9 at 13:16
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Letting $z = w - 1$, you want the Laurent expansion of $e^{w-1 \over 2 - w}$ on $|w| > 2$. Noting that ${w -1 \over 2 - w} = -1 - {1 \over w - 2}$, you are looking for the Laurent expansion of $e^{-1}e^{-{1 \over w - 2}}$ on $|w| > 2$.

Changing variables once again, this time to $v = {1 \over w}$, you are seeking the Taylor expansion of $e^{-1}e^{v \over 2v - 1} = e^{-{1 \over 2}}e^{1 \over 4v - 2}$ on $|v| < {1 \over 2}$. So $a_0 = e^{-1}$ and you can find $a_1$ and $a_2$ by taking the first two derivatives of $e^{-{1 \over 2}}e^{1 \over 4v - 2}$ at $v = 0$.

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Similar to José Carlos Santos'answer

Let $$z=x-1\implies \frac z {1-z}=\frac{1-x}{x-2}=-\frac 12 + \sum_{n=1}^\infty 2^{-(n+1)} x^n$$ which make $$e^{\frac{1-x}{x-2}}=\frac{1}{\sqrt{e}}\left(1+\frac{x}{4 }+\frac{5 x^2}{32 }+\frac{37 x^3}{384 }+\frac{361 x^4}{6144 }\right)+O\left(x^5\right)$$ The numerator of the coefficients correspond to sequence $A025168$ in $OEIS$ and the denominator of the coefficients is just $4^n n!$.

Replave $x$ by $(z+1)$.

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