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Complex numbers identity

  1. May 22, 2012 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    I'm trying to follow some solution to an exercise in physics and apparently [itex]e^{-im \frac{3\pi}{2}}=i^m[/itex] where [itex]m \in \mathbb{Z}[/itex].
    I don't realize why this is true.

    2. Relevant equations
    Euler's formula.


    3. The attempt at a solution
    I applied Euler's formula but this is still a mistery.
    [itex]i^m=\cos \left ( \frac{3\pi m}{2} \right ) -i \sin \left ( \frac{3\pi m }{2} \right )[/itex].
    I've checked the formula for m=1 and 2, it works. I must be missing the obvious, but I'm very tired physically and mentally.
    Thanks for any help.

    Edit: I found it. I drew a mental sketch of [itex]e^{-i \frac{3\pi }{2}}[/itex], it's "i" in the complex plane. Then just elevate this to the power m and the job is done.
     
  2. jcsd
  3. May 23, 2012 #2
    What you can do is to use
    [tex]i^m = e^{\ln(i^m)} = e^{m \ln i} [/tex]
    and then calculate what is [itex] \ln i [/itex]
     
  4. May 23, 2012 #3

    HallsofIvy

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    [itex]e^{im\frac{-3\pi}{2}}= \left(e^{i\frac{-3\pi}{2}}\right)^m[/itex]

    And, of course, [itex]e^{i\frac{-3\pi}{2}}= i[/itex] so that expression is just [itex]i^m[/itex].

    If you are not clear that [itex]e^{i\frac{3\pi}{2}}= i[/itex], recall that [itex]e^{i\theta}[/itex], for any real [itex]\theta[/itex], lies on the unit circle ([itex]|e^{i\theta}|= 1[/itex] at angle [itex]\theta[/itex] measured counter clockwise from the positive real axis.
    [itex]e^{i\frac{-3\pi}{2}}[/itex] lies on the unit circle, an angle [itex]3\pi/2[/itex] measured clockwise from the positive real axis.

    Another way to see this is to recall that [itex]x^{-1}= 1/x[/itex] and that [itex]1/i= -i[/itex].
     
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