Consider the Laurent expansion of $f(z)=\displaystyle{exp(\frac{z}{1-z})=\sum_{n=-\infty}^{\infty}}a_n (z+1)^n$, which converges on domain ${\{z\in \mathbb{C}:|z+1|>2}\}$. Find the coefficients $a_0$, $a_{-1}$, and $a_{-2}$.
I believe I’ve found the Laurent series for $f(z)$:
$\displaystyle{\frac{1}{e} \sum_{k=0}^{\infty} \frac{1}{k!}\Big{(} \frac{-1}{2} \sum_{n=0}^{\infty} \Big{(} \frac{2}{z+1} \Big{)}^{n+1}} \Big{)}^k$
I believe over correctly identified that $a_0= \frac{1}{e}$ and $a_{-1}=\frac{-1}{e}$. However I’m having a lot of trouble identifying $a_{-2}$. I would really like to know if there’s a way to find this coefficient without using integrals.
The function $$f(z)=\exp(z/(1-z))=\exp(-z/(z+1-2))$$ is holomorphic in the domain $|z+1|>2.$ Its Laurent expansion is of the form $$f(z)=\sum_{n=0}^\infty a_{-n}(z+1)^{-n}$$ Let $w=(z+1)^{-1}.$ Then $$\sum_{n=0}^\infty a_{-n}w^n=f(w^{-1}-1)=:g(w)$$ Hence $$a_0=g(0),\ a_{-1}=g'(0),\ a_{-2}={1\over 2}g''(0)$$ We have $$g(w)=\exp((1-w)/(2w-1))=e^{-1/2}\exp(1/(2(2w-1))$$Can you take it from here ?