# Thread: Help with Laurent Expansions

1. ## Help with Laurent Expansions

Basically I've been working through the problem below, and when I went to check my answers they differed, so I was wondering are the answers below the question both the same? Thanks.

2. Originally Posted by Monkens
Basically I've been working through the problem below, and when I went to check my answers they differed, so I was wondering are the answers below the question both the same? Thanks.

$\displaystyle z^2-3iz-2=(z-i)(z-2i)\Longrightarrow$ the function $\displaystyle f(z)=\frac{1}{z^2-3iz-2}$ is analytic in the given

annulus and thus its Laurent expansion there is just a power series...

Tonio

3. First we separate the two terms of f(*)...

$\displaystyle \displaystyle f(z)= \frac{1}{z^{2} -3\ i\ z -2} = \frac{i}{z-i} - \frac{i}{z-2\ i}$ (1)

... and then we consider that is...

$\displaystyle \displaystyle \frac{-i}{z-2\ i} = \frac{-i}{-i + (z-i)}= \frac{1}{1+i (z-i)} =$

$\displaystyle \displaystyle = \sum_{n=0}^{\infty} (-i)^{n} (z-i)^{n}$ (2)

... so that for $\displaystyle 0<|z-i|<1$ is...

$\displaystyle \displaystyle \frac{1}{z^{2} -3\ i\ z -2} = \frac{i}{z-i} + \sum_{n=0}^{\infty} (-i)^{n} (z-i)^{n}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$