Radius of Convergence for $f(z)=e^z / (z+1)^2 $

Question

I'm working on the following question in preparation for my exam.

Derive the Laurent series of $$ f(z)=\frac{e^z}{(z+1)^2} $$ around $z=-1$. What is the radius of convergence?

I got the Laurent series easy enough:

For $$ a_n = \frac{f^{(n)}(-1)}{n!} $$ Let $$ f(z)=e^z \Rightarrow f^{(n)}(z)=e^z \Rightarrow a_n =\frac{1}{e n!} \\ \Rightarrow e^z= \sum_{n=0}^\infty \frac{(z+1)^n}{e n!} $$

So, $$ \frac{e^z}{(z+1)^2} = (z+1)^{-2} \sum_{n=0}^\infty \frac{(z+1)^n}{e n!}=\sum_{n=0}^\infty \frac{(z+1)^{n-2}}{e n!} $$

I'm generally confused on how to determine the radius of convergence for complex series. Is the following work correct? If not, why?

Using the ratio test, $$ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{1}{e(n+1)!}}{\frac{1}{en!}} \right| = \left| \frac{n!}{(n+1)!} \right| \\ \Rightarrow \lim_{n\rightarrow \infty} \frac{n!}{(n+1)!} = 0$$

Therefore, the radius of convergence is $0$, implying that the series converges to a single point. Is this point $z=-1$?

Thanks!

Answer 1

A Laurent series in a variable $w$ typically converges on an annulus $\{w:r<|w|<R\}$. In this case, $w=z+1$, and the Laurent series converges everywhere save for $w=0$: in this case then $r=0$ and $R=\infty$.

Answer 2

Since $e^w=\sum_{n\geq 0}\frac{w^n}{n!}$ is an entire function we have $$ \frac{e^{w-1}}{w^2} = \frac{1}{ew^2}+\frac{1}{ew}+\color{blue}{\frac{1}{e}\sum_{n\geq 0}\frac{w^n}{(n+2)!}} $$ in a punctured neighbourhood of $w=0$. Replace $w$ with $z+1$ and you are done, you do not really need to compute any derivative: the blue part is an entire function.