Find Laurent series at $z_0 = 1$

Question

Find the Laurent series that converges for $0 < |z - z_0| < R$ and determine the precise region of convergence. Show details

$\frac{e^z}{(z-1)^2}$ around $z_0 = 1$

So if I'm going to compare this to another well-known Taylor series, what could I do? It looks like the geometric series but there's a power. Can someone show me the way?

Answer 1

You can start with the Taylor series of $e^z$ at $z_0 = 1$: $$ e^z = \sum_{n=0}^\infty \frac{e}{n!} (z-1)^n = e + e(z-1) + \sum_{n=2}^\infty \frac{e}{n!} (z-1)^n $$

Dividing this by $(z-1)^2$ leads to the Laurent series of $\frac{e^z}{(z-1)^2}$ at $z_0 = 1$: $$ \frac{e^z}{(z-1)^2} = \frac{e}{(z-1)^2} + \frac{e}{z-1} + \sum_{n=0}^\infty \frac{e}{(n+2)!} (z-1)^n $$

To calculate the radius $R$, let $a_n := \frac{e}{(n+2)!}$. Then $$ \frac 1 R = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{1}{n+3} = 0 \; , $$ so $R = \infty$.