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Engineering Math - Complex Roots and Powers

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following in x+iy form.

    2. Relevant equations

    3. The attempt at a solution
    i3+i = i3*ii

    I understand how it all works out except [itex]\frac{pi}{2}[/itex]. I can't figure out how they got [itex]\frac{pi}{2}[/itex] in the first place. Any ideas? Thanks in advance.
  2. jcsd
  3. Sep 30, 2011 #2


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    Staff: Mentor

    My editor won't allow greek letters, so in place of theta I'll use T

    You know: eiT=cosT + i.sinT

    So let T=Pi/2, so that cosT=0, and we have

    ei.Pi/2= i

    Q: How do you say in words: rei.theta

    I'm wondering what the r is part of?
  4. Sep 30, 2011 #3


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    Science Advisor

    If [itex]y= i^i[/itex], then [itex]ln(y)= i ln(i)[/itex]. Further, i has modulus r= 1 and argument [itex]\theta= \pi/2[/itex] so that [itex]i= e^{i(\pi/2+ 2k\pi)}[/itex] for k any integer ([itex]e^{2k\pi}= 1[/itex] for all integer k so these are all different ways of writing i). Then [itex]ln(i)= i(\pi/2+ 2k\pi)[/itex] and so [itex]ln(y)= -(\pi/2+ 2k\pi)[/itex].
  5. Sep 30, 2011 #4
    The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
    As far as how r is used, I'm unsure. None of the problems I have been given/done have used that r at all.
  6. Sep 30, 2011 #5
    Thank you both for the explanations! You're my heroes.
  7. Sep 30, 2011 #6


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    Staff: Mentor

    I guessed that r was the magnitude, but you confused me by introducing it in (3) for i= and I know i has a magnitude of unity. The square root of (cos2 + sin2) is always 1.

    I just wanted to make sure "re" wasn't a combination that I am unfamiliar with.

    Well, 5.cosT + 5.i.sinT would equal 5.eiT so that's when it's needed.
  8. Sep 30, 2011 #7
    Yeah, I think you're probably right. Thanks again! :)
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