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Finding Region of convergence for complex series

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    I am struggling to answer this question please help

    Find the region of convergence for the following complex series and draw the region

    Ʃ(i+z)^(2n-1)/2^(2n+1)


    3. The attempt at a solution

    Here is my hand written working, sorry i could figure out how to use the symbols

    http://i43.tinypic.com/10xsnrc.jpg

    Thank you


    I am sorry I just remembered that links to images are not permitted. How do I delete this thread?
     
    Last edited: Apr 9, 2012
  2. jcsd
  3. Apr 9, 2012 #2

    Dick

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    I think the figure is all right. Now you just want to figure out what region |(z+i)^2/4|<1 corresponds to. Simplify it a little.
     
  4. Apr 9, 2012 #3
    Thats the problem mate. I just don't have a clue on what to do after this stage. I have look a various text books and online without success.
     
  5. Apr 9, 2012 #4

    micromass

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    You want to get it in the form [itex]|z-c|<r[/itex] for some c and r.

    Hint:

    1) |cz|=c|z| for positive real c
    2) |z²|=|z|²
     
  6. Apr 9, 2012 #5
    To be honest I dont have a clue now as in most cases we have only worked out if the series in convergence or divergent. This is new to me. Are they any textbooks or websites u recommend for learning this stuff?

    Anyway thanx for your help guys
     
  7. Apr 9, 2012 #6

    Dick

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    You may be thinking this is much harder than it actually is. You've already done the hard part. Do you know |z+i|=|z-(-i)|<r is an open disk of radius r around the point z=(-i)?
     
  8. Apr 10, 2012 #7
    No I dont know that. So in this case c=-i?
     
  9. Apr 10, 2012 #8

    Dick

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    Sure, |z-c|<r is a disk centered on c. So if c=(-i) then you get |z+i|<r. Now what's r?
     
  10. Apr 10, 2012 #9
    Since its convergence r is equals to 1. Unless i have to multiply by the denominator of 4?
     
  11. Apr 10, 2012 #10

    Dick

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    Work it out. If |(z+i)^2/4|<1 then |(z-(-i)|<? No, it's not 1.
     
  12. Apr 10, 2012 #11
    cross multiplication makes it |(z+i)^2|<4
    then square root of 4 =2
    so |(z-(-i)|< 2
     
  13. Apr 10, 2012 #12

    Dick

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    Yes. The radius of convergence is 2.
     
  14. Apr 10, 2012 #13
    Oh thank you very much I have learnt something very important. So now I just need to plot xy plane with i along the y axis and numbers (radius) along the x axis. The circle with initiate from point -i with radius of 2. So how will i know that the radius is 2 along the y axis since its a imaginary axis with no real numbers?
     
  15. Apr 10, 2012 #14

    Dick

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    You are making this sound more complicated than it is. If y is your imaginary axis then the point -i is the point (0,-1). Use that as a center and draw a circle of radius 2 around it.
     
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