Help with Laurent Expansions

**Monkens** · January 16th 2011, 12:12 PM

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.

**tonio** · January 16th 2011, 05:56 PM

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.

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

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

Tonio

**chisigma** · January 17th 2011, 05:45 AM

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

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

Thread: Help with Laurent Expansions

LinkBack

Thread Tools

Display

Help with Laurent Expansions

Similar Math Help Forum Discussions

Laurent Expansions

Expansions

Laurent expansions

Laurent expansions

Complex Analysis - Cauchy Riemann Equations + Laurent Expansions

Search Tags