# Thread: Laurent series of a function

1. ## Laurent series of a function

I must find the Laurent series of the function $g(z)=\frac{1}{z(z^2+1)}$ for 3 different regions: $0<|z-i|<1$, $1<|z-i|<2$ and $2<|z-i|<\infty$.
My attempt: I've found out that g has 3 singularities (at $z=\pm i$ and $z=0$) and is I think otherwise analytic.
So for the first region, there is 1 singularity inside it (just at the center of the region).
So I think I should write the Laurent series centered at the singularity $z=i$.
I've tried many things like rewriting $g(z)=\frac{1}{z(z+i)} \cdot \frac{1}{z-i}$ but I realized it was better to keep it as it was. I was thinking about writing $\frac{1}{z^2+1}$ as an infinite series (now I realize that as it should be centered in i, I should have factored out by (z-i)...) and then divide by z the whole series. I'm going into circles and I don't click yet on how to get it. I'd like a tip.

2. A possible confortable way is the complex variable subsitution $p= \frac{1}{z-i}$ so that is...

$\displaystyle \frac{1}{z} = \frac{p}{1+i p}$

$\displaystyle \frac{1}{z+i} = \frac{p}{1+2 i p}$ (1)

... and we have...

$\displaystyle \frac{1}{z}\ \frac{1}{z-i}\ \frac{1}{z+i} = \frac{p^{3}}{(1 + i p)\ (1 + 2 i p)} = p^{3}\ (\frac{1}{1 + 2 i p} - \frac{1}{1 + i p})$ (2)

Now for $|p|<\frac{1}{2}$ is...

$\displaystyle \frac{1}{1 + i p} = 1 - i p - p^{2} + i p^{3} + ...$

$\displaystyle \frac{1}{1 + 2 i p} = 1 - 2 i p - 4 p^{2} + 8 i p^{3} + ...$ (3)

... so that is...

$\displaystyle \frac{p^{3}}{(1 + i p)\ (1 + 2 i p)} = -i p^{4} -3 p^{5} + 7 i p^{6} + ...$ (4)

... and therefore...

$\displaystyle \frac{1}{z\ (1+z^{2})} = - \frac{i}{(z-i)^{4}} - \frac{3}{(z-i)^{5}} + \frac{7 i}{(z-i)^{6}} + ...$ (5)

The series (5) converges for $|z-i|>2$ ...

Kind regards

$\chi$ $\sigma$

3. Thank you. That was actually very hard. So the main idea was to make a change of variable. Also, how do you reach that the series (5) converges for |z-i|>2?
Edit: Oh nevermind I get it! Because of line 6 of your post. Great, many thanks.