# Laurent series

1. Apr 17, 2017

### cragar

1. The problem statement, all variables and given/known data
expand $f(z)=\frac{1}{z(z-1)}$ in a laurent series valid for the given annular domain.
|z|> 3

2. Relevant equations

3. The attempt at a solution
first I do partial fractions to get
$\frac{-1}{3z} +\frac{1}{3(z-3)}$
then in the second fraction I factor out a z in the denominator to give me a geometric series
where 3/z is the common ration and converges for 3/z<1 and then 3<z.

so I get for my series $\frac{-1}{3z}+\frac{1}{3z}[1+\frac{3}{z}+\frac{3^2}{z^2}+..... ]$

2. Apr 17, 2017

### Buzz Bloom

Hi cragar:

When I combine -1/3z + 1/3(z-3) I get 1/z(z-3).

Regards,
Buzz

3. Apr 17, 2017

### cragar

oh right i mistyped the problem it has a z-3 factor in the original factor

4. Apr 17, 2017

### Buzz Bloom

Hi cragar:

The series simplifies to 3/z + 32/z2 + 33/z3 ...
The problem statement requests a series that converges for |z| > 3. Be careful to keep the absolute value notation.The region of convergence consists of all complex numbers outside the circle of radius 3 with its center at the origin z=0.

Regards,
Buzz

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