# Homework Help: Laurent series around infinity

1. Oct 14, 2013

### SqueeSpleen

Laurent series at infinity point

I already calculated it, but my work was too long, I really wish to find a shorter route.
Calculate the Laurent series of $\frac{1}{(z^{2}+1)^{2}}$ around $z_{0} = 0$
First, I used simple fractions and I got:
$\frac{1}{(z^{2}+1)^{2}}=\frac{-i}{4} \frac{1}{z-i} + \frac{i}{4} \frac{1}{z+i} + \frac{-1}{4} \frac{1}{(z-i)^{2}} + \frac{-1}{4} \frac{1}{(z+i)^{2}}$
Then I did:
$\frac{1}{z-i}=\frac{1}{-i(1+iz)}= i \displaystyle \sum_{n=0}^{\infty} (-iz)^{n}$
$\frac{1}{z+i}=\frac{1}{i(1-iz)}= -i \displaystyle \sum_{n=0}^{\infty} (iz)^{n}$
$\frac{1}{(z-i)^{2}} = \frac{\partial}{\partial z} ( - \frac{1}{z-i} )$
$\frac{1}{(z+i)^{2}} = \frac{\partial}{\partial z} ( - \frac{1}{z+i} )$
Then we got:
$\frac{1}{(z-i)^{2}} = \frac{\partial}{\partial z}( -i \displaystyle \sum_{n=0}^{\infty} (-iz)^{n} ) = -i \displaystyle \sum_{n=0}^{\infty} (n+1) (-i) (-iz)^{n}$
$\frac{1}{(z+i)^{2}} = \frac{\partial}{\partial z}( i \displaystyle \sum_{n=0}^{\infty} (iz)^{n} ) = i \displaystyle \sum_{n=0}^{\infty} (n+1) (i) (iz)^{n}$
Then we have
$\frac{1}{4} (-i i \displaystyle \sum_{n=0}^{\infty} (-iz)^{n} + i(-i) \displaystyle \sum_{n=0}^{\infty} (iz)^{n} + (-1) (-i) \displaystyle \sum_{n=0}^{\infty} (n+1) (-i) (-iz)^{n} + (-1) i \displaystyle \sum_{n=0}^{\infty} (n+1) (i) (iz)^{n} =$
$= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} (-iz)^{n} + \displaystyle \sum_{n=0}^{\infty} (iz)^{n} + i \displaystyle \sum_{n=0}^{\infty} (n+1) (-i) (-iz)^{n} + (-i) \displaystyle \sum_{n=0}^{\infty} (n+1) (i) (iz)^{n} ) =$
$= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} (-iz)^{n} + (iz)^{n} + i(n+1) (-i) (-iz)^{n} + (-i) (n+1) (i) (iz)^{n} ) =$
$= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} ((-iz)^{n} + (iz)^{n}) + (n+1)((-iz)^{n} + (iz)^{n}) )) =$
$= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} (n+2)((-iz)^{n} + (iz)^{n}))$
So we have $a_{n}=0$ $n=2k+1$ and if $n=2k$ $a_{n} = (2n+4)/4 = 2k+1$
This is the Taylor series around $z_{0}=0$
We want the Laurent series around $z_{0} = \infty$ so we do:
$\frac{1}{(z^{2}+1)^{2}} = \frac{1}{z^{4}} \frac{1}{(z^{-2}+1)^{2}}$
We take $w = \frac{1}{z}$ and when $w \to 0$ we have $\frac{1}{z} \to 0$
As $\frac{1}{z^{4}} \frac{1}{(z^{-2}+1)^{2}} = w^{4} \frac{1}{(w^{2}+1)^{2}}$, the Laurent series of this function is the same we previously calculated with a +4 in the exponent.
We revert the change of variable and we got the series.

$\frac{1}{4} \displaystyle \sum_{n=0}^{\infty} (n+2)((\frac{1}{iz})^{n+4} + (\frac{1}{-iz})^{n+4})$

Finally, there are a lots of things I did I'm not really sure if were well done, I verified my work with calculator and later with wolframalpha, but I'm not sure what I this is right.

Last edited: Oct 14, 2013
2. Oct 14, 2013

### jackmell

That's well-done in my opinion. Suppose you could have outright made the substitution $z\to 1/w$ to obtain

$$w^4\left(-\frac{1}{2w}\frac{d}{dw} \frac{1}{1+w^2}\right)$$

and just expand the $\frac{1}{1+w^2}$ about zero.

3. Oct 14, 2013

### SqueeSpleen

Thanks, it's a waaay shorter.
$z = \frac{1}{w} \Rightarrow \frac{1}{(z^{2}+1)^{2}} = \frac{1}{(\frac{1}{w^{2}}+1)^{2}} = \frac{1}{\frac{1}{w^{4}} ( 1+w^{2})^{2}} = w^{4} \frac{1}{(w^{2}+1)^{2}}$
$\frac{d}{d w} (\frac{1}{w^{2}+1}) = \frac{-2w}{(w^{2}+1)^{2}} \Rightarrow \frac{1}{(w^{2}+1)^{2}} = -\frac{1}{2w} \frac{d}{d w} (\frac{1}{w^{2}+1})$
$w^{4} (-\frac{1}{2w} \frac{d}{d w} (\frac{1}{w^{2}+1})) = (-\frac{w^{3}}{2} \frac{d}{d w} (\frac{1}{w^{2}+1}))$
$\frac{1}{1+w^{2}} = \displaystyle \sum_{n=0}^{\infty} (-w^{2})^{n}$
$\frac{d}{d w} \frac{1}{1+w^{2}} = \displaystyle \sum_{n=0}^{\infty} (n+1) (-w^{2})^{n} (-2w) \Rightarrow (-\frac{w^{3}}{2} \frac{d}{d w} (\frac{1}{w^{2}+1})) = \displaystyle \sum_{n=0}^{\infty} (n+1) (-w^{2})^{n} (-2w) (-\frac{w^{3}}{2})=$
$= \displaystyle \sum_{n=0}^{\infty} (n+1) (-w^{2})^{n} (w^{4}) = \displaystyle \sum_{n=0}^{\infty} (n+1) (-1)^{n} w^{2n+4} = \displaystyle \sum _{n=0}^{\infty} (n+1) (-1)^{n} \frac{1}{z^{2n+4}}$

Which is valid for all $z \in \mathbb{C} / | z | > 1$ (In the mess I forget to mention it my the previous post). We know it's valid here because it's where the geometric series equality is true.

Last edited: Oct 14, 2013