Laurent series

  • Thread starter krindik
  • Start date
  • #1
krindik
65
1
Hi,
can u pls help me on this?

Homework Statement


Find Laurent series that converges for

[tex]
\, 0 < |z - z_0| < R }
[/tex] and determine precise region of convergance

[tex]
\, \frac {1}{z^2 + 1} \,\,

[/tex]

Homework Equations





The Attempt at a Solution


I tried to spilt this into fractions
i.e
[tex]
f(x) \, = \, \frac{A}{z-i} + \, \frac{B}{z+i}

[/tex]

as I would have done for

[tex]
\frac {1}{z^2 - 1} \,\,

[/tex]

But in that case I would expand it with a geometrical series.
The problem rises with [tex] i [/tex] instead of [tex] 1 [/tex]



2. Homework Statement
Can u pls explain how can I choose the method of expansion (Laurent, Taylor) given a function f(x) ?






Thanks

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
latentcorpse
1,444
0
can't you just say [itex]w=z^2[/itex] then you can make

[itex]\frac{1}{w-1}=\frac{1}{w} \frac{1}{1-\frac{1}{w}}[/itex] and expand as a geometric series...
 
  • #3
krindik
65
1
Thanks.

In fact the question is
[tex]

\, \frac {1}{z^2 + 1} \,\,

[/tex]

not

[tex]

\, \frac {1}{z^2 - 1} \,\,

[/tex]




I'll restate the question

Find Laurent series that converges for

[tex]

\, 0 < |z - z_0| < R }
[/tex]

and determine precise region of convergance for
[tex]

\, \frac {1}{z^2 + 1} \,\,

[/tex]

where
[tex]
z_0 = i
[/tex]



Additionally, I would like to know how u use [tex]

\, 0 < |z - z_0| < R }
[/tex]
in coming up with a solution
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
26,263
621
The Laurent series will be an expansion in powers of (z-i). Use your partial fractions form. The A/(z-i) is the (-1) power. Write the B/(z+i) part as B/(2i+(z-i)) and do the geometric series expansion with (z-i) as the variable.
 
  • #5
latentcorpse
1,444
0
why can we not just change variable here?
 
  • #6
Dick
Science Advisor
Homework Helper
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You can. Isn't that basically what I did when I wrote z+i=(z-i)+2i?
 
  • #7
krindik
65
1
Thank you,

I have few questions
1. Is it that you choose [itex] (z - i) [/itex] terms because the question specifies it should converge in [itex] 0 < |z-i| < R [/itex] ?

2. So [itex]\frac{B}{z + i} [/itex] will be expanded like

[itex]\frac{1}{2i} * \frac{1}{1+\frac{z-i}{2i}} = \sum_{n=0}^{\infty}{s^n}[/itex]

where
[itex] s = \frac{z-i}{2i}[/itex]


so the answer to the solution would be
[itex]\frac{A}{z-i} + \{expansion\, in\, 2\}[/itex] ?

Thanks again
 
  • #8
Dick
Science Advisor
Homework Helper
26,263
621
I chose (z-i) because you specified that you want to expand around z0=i in post 3. I couldn't tell where you wanted to put the expansion from the first post. And yes, you've got the expansion right.
 
  • #9
krindik
65
1
Thank you very much Dick

I shall summarize here, so that anybody else following this thread can get the point.


[itex]
\frac{1}{1+z^2} = \frac{A}{z-i} + \frac{B}{z+i}
[/itex]

expand
[itex]
\frac{B}{z+i} = \frac{B}{2i} * \frac{1}{1-(-1 * \frac{z-i}{2i})}
[/itex]

then by expanding in geometric series
[itex]
\frac{1}{1-(-1 * \frac{z-i}{2i})} = \sum_{n=0}^\infty{[(\frac{-1}{2i})^n * (z-i)^n]}
[/itex]
 

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