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Complex Solutions

  1. Jul 10, 2008 #1
    Hi,
    Could anyone please explain to me the procedure of obtaining complex solutions of equations where i is involved? For example z^3=i.
    Many thanks!
     
  2. jcsd
  3. Jul 10, 2008 #2
    -i is one solution. I obtained that via more madness than method
     
  4. Jul 10, 2008 #3

    nicksauce

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    The usual thing to do would be to write the equal in exponential notation, ie [itex]z = re^{i\theta}}[/itex], [itex]i=e^{i\frac{\pi}{2}}[/itex], then for z^3 = i, we would have
    [tex]r^3e^{3i\theta} = e^{i\frac{\pi}{2}}[/tex]
    From this you can conclude that r = 1. To find theta you take the logarithms,
    [tex]3i\theta + 2ik\pi = i\frac{\pi}{2}[/tex]
    And then solve for theta, with different values of k (k as an integer).
     
  5. Jul 10, 2008 #4
    You do it the same way you find the complex solutions of e.g. z^3=1
     
  6. Jul 10, 2008 #5
    thanks a bunch nicksauce :)
     
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