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.