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!

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- Aug 25th 2010, 06:04 AMschrodingersdogLaurent 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, 06:57 AMBruno J.
Use partial fractions and then geometric series.

- Aug 25th 2010, 10:43 AMschrodingersdog
how do you partial fractions when there's a complex number in the denominator?

- Aug 25th 2010, 05:13 PMBruno J.
- Aug 26th 2010, 08:18 AMchisigma
For (i) You can use the substitution $\displaystyle s= \frac{1}{1+z}$ that with some steps transforms the function...

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

... into ...

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

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

$\displaystyle \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...

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

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

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

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

Kind regards

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