How to Expand a Complex Function into a Laurent Series?

[aruwin]

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.

I have no idea how to start.
I guess geometric series could be applied later but I don't know how to start.

 

[chisigma]

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.

I have no idea how to start.
I guess geometric series could be applied later but I don't know how to start.

Your start-up is very good!... You can proceed as follows...

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

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

Kind regards

$\chi$ $\sigma$

 

[aruwin]

$\displaystyle = - \frac{\frac{1}{2}}{z+i} - \frac{1}{4\ i}\ \frac{1}{1 - \frac{z+i}{2 \ i}} $
Kind regards

How did you change the equation into this?

 

[chisigma]

How did you change the equation into this?

... is ...

$\displaystyle \frac{\frac{1}{2}}{z-i} = \frac{\frac{1}{2}}{z + i - 2\ i}= - \frac{\frac{1}{4\ i}} {1 - \frac{z+i}{2 \ i}} $

Kind regards

$\chi$ $\sigma$

 

[aruwin]

... is ...

$\displaystyle \frac{\frac{1}{2}}{z-i} = \frac{\frac{1}{2}}{z + i - 2\ i}= - \frac{\frac{1}{4\ i}} {1 - \frac{z+i}{2 \ i}} $

Kind regards

$\chi$ $\sigma$

Oh, you divide everything by 2i, right?

 

[aruwin]

Your start-up is very good!... You can proceed as follows...

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

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

Kind regards

$\chi$ $\sigma$

I re-write the function this way, 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 ∑ in the final answer. How to do that?