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Laurent series part 2

  1. Jul 23, 2014 #1
    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.
     
  2. jcsd
  3. Jul 23, 2014 #2

    vanhees71

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    [itex]z-\mathrm{i}=z+\mathrm{i}-2 \mathrm{i}[/itex]...
     
  4. Jul 23, 2014 #3
    What is the advantage of writing it like that?
     
  5. Jul 23, 2014 #4

    HallsofIvy

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    Write your fraction as [itex]-\frac{1}{z+ i}\frac{1}{2i- (z+ i)}[/itex] and expand the second fraction as a geometric series in z+i.
     
  6. Jul 23, 2014 #5
    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?
     

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  7. Jul 23, 2014 #6
    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?
     
  8. Jul 23, 2014 #7

    vanhees71

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    Just multiply the factor in the front into your sum, and you are done :-)).
     
  9. Jul 23, 2014 #8
    Thank you! This is solved :)
     
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