<s>[INST] Laurent Expansion Coefficients of $\exp\left(\frac{z}{1-z}\right)$ 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.
 [/INST] 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 $(z+1)$.
</s>