# Find Laurent series at $z_0 = 1$

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?

• $e^{z}=e \sum \frac {(z-1)^{n}} {n!}$ Nov 26, 2020 at 6:42

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 $$R = \infty$$.