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