Laurent series

1. May 28, 2010

squenshl

1. The problem statement, all variables and given/known data
Calculate a Laurent series about z = 0 for 1/(z2(z+i)) in the region D = {z: |z| < 1}

2. Relevant equations

3. The attempt at a solution
I used partial fractions to get 1/(z2(z+i)) = 1/z -1/z2 - 1/(z+i) but where do I go from here.

2. May 29, 2010

HallsofIvy

Staff Emeritus
Expand $$\frac{1}{z+i}$$ in a McLaurin series (Taylor's series about z= 0) and multiply each term by $z^{-2}$.

You can get the McLaurin series most easily by writing $$\frac{1}{z+i}= -i\frac{1}{1- (-z/i)}$$$$=-i\frac{1}{1- (iz)}$$ and writing it as a geometric series.

3. May 29, 2010

squenshl

I let t = z
So 1/(t+i) = 1/t(1+i/t) which is a geometric series
= 1/t(1-(i/t)+(-1/t)^2-...
= 1/t - i/t2 + i2/t3 + ...
= 1/z - i/z2 - 1/z3 + ...

Last edited: May 30, 2010
4. May 30, 2010

squenshl

How do I find the geometric series for 1/z and -i/z2

5. May 30, 2010

vela

Staff Emeritus
That series doesn't converge for |z|<1.

You don't. The 1/z and 1/z2 terms are already in the correct form.

6. May 30, 2010

Cheers.