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Homework Help: How to find all complex Z such that Z^5=-32

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data
    How to find all complex number Z such that Z^5=-32


    2. Relevant equations
    Euler equation
    e^it=cost+isint


    3. The attempt at a solution
    I guess a naive way to solve is that since Z^5=(-2)^5
    Therefore Z=-2, but this obviously too good to be true.
    I have no background in complex analysis.
     
  2. jcsd
  3. Sep 8, 2010 #2

    Dick

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    Z=(-2) isn't too good to be true. You are right. It is a root. But there are four more. Express -32 in polar form. I.e. -32=32*e^(it). The use deMoivre. What are the possibilities for t? There are more than one. Give me one to start out with.
     
    Last edited: Sep 8, 2010
  4. Sep 8, 2010 #3
    Thank you for your reply.
    Sorry I do not understand that -32=32*e^(it)?
    e^it=cost+isint
    I mean t is a variable that takes on arbitrary values.
     
  5. Sep 8, 2010 #4

    Dick

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    t isn't arbitrary if -32=32*e^(it). That means e^(it)=(-1). What's t?
     
  6. Sep 8, 2010 #5
    t=180degree=-5pi,-3pi,-pi,pi,3pi,5pi,7pi,9pi,11pi?
    Therefore -32=32*e^(it)=32*[cos(pi)+isin[pi]]

    Sorry how do I use de Moivre's formula? Never taken complex analysis
     
  7. Sep 8, 2010 #6

    Dick

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    You could look it up rather than me telling you. You multiply the arguments (the angles) by 1/5. How many roots do you get before they start repeating?
     
    Last edited: Sep 8, 2010
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