Expanding f(z) in a Laurent Series for |z|>3

[cragar]

Homework Statement

expand $f(z)=\frac{1}{z(z-1)}$ in a laurent series valid for the given annular domain.
|z|> 3

Homework Equations

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 $\frac{-1}{3z}+\frac{1}{3z}[1+\frac{3}{z}+\frac{3^2}{z^2}+... ]$

 

[Buzz Bloom]

Hi cragar:

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

Regards,
Buzz

 

[cragar]

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

 

[Buzz Bloom]

converges for 3/z<1 and then 3<z.

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