# Thread: Laurent series

1. Expand in Laurent series about Zo= 0 :
$\displaystyle f(z) = \frac{z }{{(z+1)^2}(z-2)}$ i know there are three cases but i don't know how i starting ??
this is true or not ??:
$\displaystyle \frac{z }{{(z+1)^2}(z-2)} = \frac {2}{9(z-2)} - \frac{2}{9(z+1)} + \frac {1}{3{(z+1)^2}}$

And how i can starting if $\displaystyle f(z) =Log (1+ \frac{1}{z-1} )$ near Zo=1

2. Can you find the Laurent series of $\displaystyle \frac{1}{1+z}$ about $\displaystyle z=0$? Then you can find the Laurent series of $\displaystyle \frac{1}{(1+z)^2}$ by differentiating.

To find the Laurent series of $\displaystyle \frac{1}{z-2}$ about $\displaystyle z=0$ you can use the identity $\displaystyle \frac{1}{z-2}=\frac{-1}{2}\frac{1}{1-\frac{z}{2}}$.

Then you can just multiply the series (which will work since you will not have any negative powers of $\displaystyle z$ in any of your answers).

3. Originally Posted by Bruno J.
To find the Laurent series of $\displaystyle \frac{1}{z-2}$ about $\displaystyle z=0$ you can use the identity $\displaystyle \frac{1}{z-2}=\frac{-1}{2}\frac{1}{1-\frac{z}{2}}$.
That will give you the Laurent series that converges when |z|<2. In the region |z|>2 you need to write it as

$\displaystyle \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}} = \frac1z\Bigl(1+\tfrac2z + \bigl(\tfrac2z)^2+\ldots\Bigr)$.

In general, the Laurent series for $\displaystyle \frac1{z-a}$ will consist of powers of $\displaystyle \tfrac za$ in the region |z|<a, and powers of $\displaystyle \tfrac az$ in the region |z|>a.

So for the function $\displaystyle \frac{z }{{(z+1)^2}(z-2)} = \frac {2}{9(z-2)} - \frac{2}{9(z+1)} + \frac {1}{3{(z+1)^2}}$ there are three separate Laurent expansions, obtained by adding the power series expansions of the partial fractions. In the region |z|<1 you need to use positive powers of z/2 for the first fraction, and positive powers of z for the other two fractions. In the region 1<|z|<2 you need to use positive powers of z/2 for the first fraction, and negative powers of z for the other two fractions. In the region |z|>2 you need to use negative powers of z/2 for the first fraction, and negative powers of z for the other two fractions.

4. Originally Posted by Opalg
That will give you the Laurent series that converges when |z|<2. In the region |z|>2 you need to write it as

$\displaystyle \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}} = \frac1z\Bigl(1+\tfrac2z + \bigl(\tfrac2z)^2+\ldots\Bigr)$.

In general, the Laurent series for $\displaystyle \frac1{z-a}$ will consist of powers of $\displaystyle \tfrac za$ in the region |z|<a, and powers of $\displaystyle \tfrac az$ in the region |z|>a.

So for the function $\displaystyle \frac{z }{{(z+1)^2}(z-2)} = \frac {2}{9(z-2)} - \frac{2}{9(z+1)} + \frac {1}{3{(z+1)^2}}$ there are three separate Laurent expansions, obtained by adding the power series expansions of the partial fractions. In the region |z|<1 you need to use positive powers of z/2 for the first fraction, and positive powers of z for the other two fractions. In the region 1<|z|<2 you need to use positive powers of z/2 for the first fraction, and negative powers of z for the other two fractions. In the region |z|>2 you need to use negative powers of z/2 for the first fraction, and negative powers of z for the other two fractions.
Opalg>>>thanks !