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Powers of Complex Numbers

  1. Jan 15, 2005 #1

    cepheid

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    Hello,

    I'm having trouble with this problem:

    [tex] \left| \frac{(\pi + i)^{100}}{(\pi - i)^{100}} \right| = \ \ ? [/tex]

    My first thought was, "put it in polar form and simpify," but that is not helping.

    For the numerator pi + i :
    [tex] r = \sqrt{\pi^2 + 1} [/tex]

    [tex] \theta = \arctan{ \frac{1}{\pi} } [/tex] = ???

    I don't see how this will help, it's not an easy one to put in polar form

    I can also see that the numerator and denominator are complex conjugates, so maybe that is the starting point. But I can't see how to proceed
     
  2. jcsd
  3. Jan 15, 2005 #2

    Curious3141

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    Hint : they're only asking for the magnitude of the ratio of the two. Since, the numerator and denominator are complex conjugates, what can you say about their magnitudes ? Don't they cancel out ? What will be left when the magnitudes cancel out ?

    This is a trick question, with a trivial solution.
     
  4. Jan 15, 2005 #3
    Sorry to break into your thread, but I suddenly have an irrepressible need to know if this is allowed in the world of complex numbers:

    [tex] \left | \frac{(pi + i)^{100}}{(pi - i)^{100}}\right | { }^?_=
    \left ( \frac{|(pi + i)|}{|(pi - i)|} \right )^{100} [/tex]

    Is it?
     
  5. Jan 15, 2005 #4

    Curious3141

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    Yes, that works out. In fact, the numerator and demoninator can be any old complex numbers, not necessarily conjugates.
     
  6. Jan 15, 2005 #5

    HallsofIvy

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    In other words, you don't NEED to know the argument. All you are asked about is the modulus so that's all you need to know!

    In general, |zn|= |z|n and |a/b|= |a|/|b|.
     
  7. Jan 15, 2005 #6

    cepheid

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    I can't believe I didn't see that! Even though I realised the numerator and denominator were conjugates.

    Thanks for your help, everyone.

    Halls:

    Yeah, we also had to prove |z1z2| = |z1||z2| in this hw assignment
    so I can see where those come from.

    Thanks again.
     
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