# Use geometric series to find the Laurent series

1. May 10, 2012

### Kiefer

Use geometric series to find the Laurent series for f (z) = z / (z - 1)(z - 2) in each annulus
(a) Ann(1,0,1)
(b) Ann(1,1,∞)

Ann(a,r,R)
Ann(1,0,1)=D(1,1)\{0}

My attempt:
f(z)= -1/(z-1) + 2/(z-2)
geometric series: Σ[n=0 to inf] z^n - 1/2 Σ[n=0 to inf] (z/2)^n

a)0<|z-1|<1
2/(z-2)=-2(1/(2-z))=-2(1/(1-(z-1)))=-2Σ[n=0 to inf](z-1)^n
f(z)=-1/(z-1)-2Σ[n=0 to inf](z-1)^n=-(z-1)^(-1)+Σ[n=0 to inf]-2(z-1)^n

b)1<|z-1|<inf
(1/|z-1|)<1
-1/(z-1)=-Σ[n=0 to inf](1/(z-1))^n
2/(z-2)=2/(z-1-1)=2(1/(z-1))(1/(1-1/(z-1)))=2/(z-1)Σ[n=0 to inf](1/(z-1))^n=2Σ[n=0 to inf](z-1)^-(n+1)
f(z)=-Σ[n=0 to inf](1/(z-1))^n+Σ[n=-inf to -2]2(z-1)^-(n+1)

Are my geometric series correct? (I feel good about the second one but am unsure on the first)
Are my summation bounds in the Laurent series correct? I did something funky in the last one, so I'm not sure. Also, I wasn't sure how to find a Laurent series for -1/(z-1) in part a, so I left it as shown.
Any help and corrections are appreciated!

2. May 11, 2012

### Dustinsfl

For $0<|z-1|<1$, I was able to only do half of it. I can't figure out the other piece unless it is supposed to be just like this.

So
$$\frac{2}{z-2} = -2\frac{1}{1+1-z}=-2\frac{1}{1-(z-1)}=-2\sum_{n=0}^{\infty}(z-1)^n$$
Then
$$f(z) = \underbrace{\frac{-1}{z-1}}_{\text{this one is giving me trouble}} - 2\sum_{n=0}^{\infty}(z-1)^n$$

For $1<|z-1|<\infty$, again I could only figure out one piece of it.

So
$$\frac{2}{z-2} = 2\frac{1}{1-\frac{1}{z-1}}=2\sum_{n=0}^{\infty}\left(\frac{1}{z-1}\right)^n$$
Then
$$f(z) = \frac{-1}{z-1}+2\sum_{n=0}^{\infty}\left(\frac{1}{z-1}\right)^n$$

I am not sure how we can contend for $\frac{-1}{z-1}$ in both a and b. Maybe it can be left as is and that is it.

3. May 11, 2012

### scurty

Part a is correct. It's okay to have your series split up like that, you don't have to have a general formula for all of your $C_n$.

Okay, part b. $\frac{-1}{z-1} \neq \displaystyle \sum_{n=0}^{\infty} \left(\frac{1}{z-1}\right)^n$

Just leave that quantity as it was in part a).

$2 \cdot \displaystyle \sum_{n=0}^{\infty} (z-1)^{-(n+1)} \neq 2 \cdot \sum_{-\infty}^{-2} (z-1)^{-(n+1)}$
$2 \cdot \displaystyle \sum_{n=0}^{\infty} (z-1)^{-(n+1)} = \sum_{-\infty}^{0} (z-1)^{n-1}$
Now you can easily change that to $(z-1)^n$.