How to Expand a Complex Function into a Laurent Series?

In summary: $\displaystyle \sum_{n=0}^{\infty}\left(\frac{z+i}{2i}\right)^{\!{n}}=\sum_{n=0}^{\infty}\left(\frac{1}{z+i}\right)^{\!{n}}=\sum_{n=0}^{\infty}\left(\frac{1}{z}\right)^{\!{n}}$
  • #1
aruwin
208
0
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.
 
Last edited:
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  • #2
aruwin said:
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

$\chi$ $\sigma$
 
  • #3
chisigma said:
$\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?
 
  • #4
aruwin said:
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

$\chi$ $\sigma$
 
  • #5
chisigma said:
... 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?
 
  • #6
chisigma said:
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?
 

Related to How to Expand a Complex Function into a Laurent Series?

1. What is the difference between a Laurent series and a Taylor series?

A Laurent series and a Taylor series are both mathematical representations of functions, but they differ in their domains. A Taylor series represents a function as a polynomial in powers of the variable, while a Laurent series represents a function as a sum of a polynomial and a series in negative powers of the variable.

2. How is a Laurent series used in complex analysis?

A Laurent series is used in complex analysis to extend the domain of a function beyond its singularities. It allows for the examination of the behavior of a function at points where it is not defined, and can help identify essential singularities or poles of a function.

3. Can a Laurent series converge on its entire domain?

No, a Laurent series can only converge on a specific annulus in the complex plane, known as the convergence region. This region is determined by the location of the singularities of the function being represented.

4. How can I find the Laurent series of a given function?

The Laurent series of a function can be found by using the formula for the coefficients of the series, which involves taking derivatives and evaluating them at the center of the series. Alternatively, software programs such as Mathematica can be used to calculate the Laurent series of a function.

5. Are there any applications of Laurent series in real-world problems?

Yes, Laurent series have many applications in physics, engineering, and other fields. For example, they are used in signal processing to approximate and analyze continuous functions. They are also used in solving differential equations and in modeling physical phenomena such as waves and oscillations.

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