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Homework Help: Absolute value squared of complex number?

  1. Apr 9, 2013 #1
    I'm given [itex]1-a\cdot e^{-i\cdot 2 \pi f}[/itex]. The squared absolute value apparently is [itex]|1-a\cdot e^{-i\cdot 2 \pi f}|^2=1+a^2-2acos(2 \pi f)[/itex].

    Sadly the awnser doesn't show the steps of this derivation. I have tried many times to derive it my self but have not been able to do so. I feel like i am missing the obvious, does any one care to show me how it's done?
  2. jcsd
  3. Apr 9, 2013 #2


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    It can be done quite easily by first expressing the exponential complex number in its trigonometric form, evaluating its absolute value and then simplifying with trigonometric identities.

    How did you approach it?
  4. Apr 9, 2013 #3

    Simon Bridge

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    Here z is the sum of two complex phasors, one length ##1## pointing along the real axis and the other length ##a## pointing at angle ##2\pi f## from the real axis. The modulus is the length of the resultant, which you find from the cosine rule.

    Algebraically you can use Euler's formula and trig identities.
  5. Apr 9, 2013 #4
    Pff i feel stupid :P

    I was trying to do all kinds of crazy stuff like interchanging the power and absolute value etc. And making use of things like [itex]|e^{i\cdot \phi}|=e^{-Im(\phi)}[/itex]. Can't believe it's that simple :P

    Thnx for the help!
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