Calculating a Laurent Series: 1/(z2(z+i))

[squenshl]

Homework Statement

Calculate a Laurent series about z = 0 for 1/(z²(z+i)) in the region D = {z: |z| < 1}

Homework Equations

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.

 

[HallsofIvy]

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.

 

[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 + ...

 

[squenshl]

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

 

[vela]

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 + ...

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

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

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

 

[squenshl]

Cheers.