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Complex exponents

  1. Apr 25, 2012 #1
    Hi, I have recently started learning about complex numbers and the complex plane, and there is something I don't quite understand. Why is it that for every number [itex] z \in C [/itex]

    [itex] e^{i\theta} = z [/itex]

    we have

    [itex] \sqrt { [Re(z)]² + [Im(z)]² } = 1 [/itex] ?

    Except for [itex] \theta = \frac {\pi}{2} or \frac {3\pi}{2} [/itex] of course. In other words, why do all numbers z define a unit circle's radius ?

    Any help would be very much appreciated
  2. jcsd
  3. Apr 25, 2012 #2
    That is not true. Only complex numbers of the form z = e^(iθ) lie on the unit circle and satisfy that equation. Complex numbers of the form z = Re^(iθ) lie on circles of radius R in the complex plane, as you can verify.
  4. Apr 25, 2012 #3
    I have stated that z must satisfy

  5. Apr 25, 2012 #4
    In that case, apply Euler's formula: e^(iθ) = cos(θ) + i*sin(θ), from which the result is immediate. The derivation of Euler's formula is implied by the Taylor series for e^x, cos(x) and sin(x), but its full justification requires a little more rigor.
  6. Apr 25, 2012 #5
    Ok so adding the taylor series for cos(θ) and for i sin(θ) should give the taylor series for e^(iθ) ? I will definitely try that out. Thank you!
  7. Apr 25, 2012 #6
    Yes, that's one of the proofs. Definitely worth going through the steps. Euler's formula falls right out of the power series for e^(iθ).
  8. Apr 25, 2012 #7


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    Also if you apply the fact that the complex conjugate of e^(iθ) = e^(-iθ) then multiply the two gives you e^(iθ - iθ) = e^(0) = 1.

    You can prove this using the propery that cos(-x) = cos(x) and -sin(x) = sin(-x) using properties of odd and even functions.
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