Laurent Series Part 2: Expanding 1/(z^2+1) at z=-i

[MissP.25_5]

Hello.

I am stuck on this question.
Let {$z\in ℂ|0<|z+i|<2$}, expand $\frac{1}{z^2+1}$ where its center $z=-i$ into Laurent series.

This is how I start off:
$$\frac{1}{(z+i)(z-i)}$$

And then I don't know what to do next. I guess geometric series could be applied later but I don't know how to rewrite this function.

 

[vanhees71]

z-\mathrm{i}=z+\mathrm{i}-2 \mathrm{i}...

 

[MissP.25_5]

z-\mathrm{i}=z+\mathrm{i}-2 \mathrm{i}...

What is the advantage of writing it like that?

 

[HallsofIvy]

Write your fraction as -\frac{1}{z+ i}\frac{1}{2i- (z+ i)} and expand the second fraction as a geometric series in z+i.

 

[MissP.25_5]

Write your fraction as -\frac{1}{z+ i}\frac{1}{2i- (z+ i)} and expand the second fraction as a geometric series in z+i.

This is what I got. I hope this is correct, but I don't know how to summarize all the terms. I need to use only one ∑. How to do that?

 

[MissP.25_5]

Is this ok?
$$\frac{1}{z+i}\times\frac{1}{(z+i)-2i}$$
$$=\frac{1}{z+i}\times[\frac{\frac{1}{(-2i)}}{\frac{-(z+i)}{2i}+1}]$$
$$=\frac{1}{z+i}\times[\frac{1}{(-2i)}\times\frac{1}{1-\frac{(2+i)}{2i}}]$$

$$=\frac{1}{z+i}\times\frac{1}{(-2i)}\times\sum_{n=0}^{\infty}\left(\frac{z+i}{2i}\right)^{\!{n}}$$

I don't know how to summarize all the terms. I need to use only one ∑. How to do that?

 

[vanhees71]

Just multiply the factor in the front into your sum, and you are done :-)).

 

[MissP.25_5]

Thank you! This is solved :)