CVS4

Find the Laurent series about $z_0=1\,$ for the function $\frac{e^{2z}}{(z-1)^3}\,$

The only singularity is at $z=1\,$ so the series will look like

$\sum_{n=-\infty}^{\infty} A_n (z-1)^n, 0<|z-1|<\infty\,$

$A_n = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-1)^{n+1}} dz = \frac{1}{2\pi i}\oint_C \frac{e^{2z}}{(z-1)^{n+4}} dz, n=0,\pm 1,\pm 2, ...\,$

where $C\,$ is any simple closed curve around the point $z=1\,$ .

Cauchy's theorem gives $A_n=0, n=-4,-5,...\,$

For $n\ge 3\,$

$A_n = \frac{1}{(n+3)!} \frac{d^{n+3}}{dz^{n+3}} e^{2z} \Bigg|_{z=1} = \frac{e^2 2^{n+3}}{(n+3)!}, n\ge 3\,$ .

The Laurent series is

$f(z) = \frac{e^{2z}}{(z-1)^3} = 8e^2\sum_{n=-3}^\infty \frac{2^n}{(n+3)!}(z-1)^n, 0<|z-1|<\infty\,$ .