Laurent series around infinity

[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.

 

[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.

 

[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.