# Laurent expansion

• Aug 25th 2010, 07:04 AM
schrodingersdog
Laurent expansion
Hey, I'm having a lot of trouble with this Laurent expansion because it's a bit more complicated than the ones I'm used to. Sorry it's not in LaTeX.

f(z) = z / (z - 2i)(z + 1)

valid for (i) 0 < |z + 1| < sqrt(5)
(ii) |z| > 2

Thanks for any help!
• Aug 25th 2010, 07:57 AM
Bruno J.
Use partial fractions and then geometric series.
• Aug 25th 2010, 11:43 AM
schrodingersdog
how do you partial fractions when there's a complex number in the denominator?
• Aug 25th 2010, 06:13 PM
Bruno J.
Quote:

Originally Posted by schrodingersdog
how do you partial fractions when there's a complex number in the denominator?

You do just as you would if it were a real number! Give it a try, and post your calculations if you get stuck.
• Aug 26th 2010, 09:18 AM
chisigma
For (i) You can use the substitution $s= \frac{1}{1+z}$ that with some steps transforms the function...

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

... into ...

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

Now if $|s|<\frac{1}{|1+2i|} = \frac{1}{\sqrt{5}}$ is...

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

... so that if You turn back to the z variable and consider that is...

$\displaystyle f(s)= \frac {z}{(1+ z)^{2}} = \frac{1}{1+z} - \frac{1}{(1+z)^{2}}$ (4)

... from (3) with some other steps you obtain finally the Laurent expansion...

$\displaystyle f(z)= \frac{1}{1+z} + \frac{2 i}{(1+z)^{2}} + \frac{2i -4}{(1+z)^{3}} + ...$ (5)

For (ii) You set $s=\frac{1}{z}$ and proceed is similar way... the computation is left to You...

Kind regards

$\chi$ $\sigma$