# Laurent series expansion

• December 2nd 2009, 09:08 PM
eskimo343
Laurent series expansion
Write the Laurent series expansion for the following functions $f(z)$ centered at $z_0$. State what are the possible domains.

(a) $f(z) = \frac{z}{5z+1}$ $z_0=1$

(b) $g(z) = \frac{z}{z^2-3iz-2}$ $z_0=0$

I know how to do part (a) when $z_0=0$ but I do not see how to do it when $z_0=1$. For part (b), I used the quadratic formula, then I did not know how to finish the problem. I need a push in the right direction. Thanks in advance.
• December 3rd 2009, 01:02 AM
Mauritzvdworm
Both problems depend on the region you will be working with, let us fist consider the first problem:

We consider the region $\|z\|>\frac{1}{5}$
$\frac{zz_{0}}{5z+1}=\frac{1}{z}\frac{z_{0}}{\frac{ 5}{z}+\frac{1}{z^{2}}}=\frac{z_{0}}{5+\frac{1}{z}} =\frac{z_{0}}{5}\frac{1}{1-\left(-\frac{1}{5z}\right)}=\frac{z_{0}}{5}\sum^{\infty}_ {j=0}\left(\frac{1}{5z}\right)^{j}$

next consider the region $\|z\|<\frac{1}{5}$
$\frac{zz_{0}}{5z+1}=zz_{0}\frac{1}{1-\left(-5z\right)}=zz_{0}\sum^{\infty}_{j=0}\left(-5\right)^{j}z^{j}=z_{0}\sum^{\infty}_{j=0}\left(-5\right)^{j}z^{j+1}$

The second problem can be done in a similar fashion, just consider the regions of interest separately, namely
i. $\|z\|<1$
ii. $1<\|z\|<2$
iii. $\|z\|>2$

you can then also make use of the following decomposition
$\frac{zz_{0}}{z^{2}-3iz-2}=\frac{zz_{0}}{\left(z-2i\right)\left(z-i\right)}=zz_{0}\left(\frac{1}{i\left(z-2i\right)}-\frac{1}{i\left(z-i\right)}\right)$

Now you will easily be able to find the solutions
• December 3rd 2009, 11:30 AM
eskimo343
I am stuck on part (b). So far I have:

$\frac{z}{z^2-3iz-2}$

now consider the regions $|z|<1$, $1<|z|<2$, and $|z|>2$. Now,

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

a) $\frac{1}{i(z-2i)}=\frac{-1}{2}(\frac{1}{1-\frac{iz}{2}})$

b) $\frac{1}{i(z-i)}=\frac{1}{1+iz}$

Now how do we get a Laurent series from all this work? I don't see how to use these intervals to get a Laurent series.