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Re[ (a+bi)^p]

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate [itex] Re[(a+bi)^p] [/itex]

    3. The attempt at a solution

    [itex](a+bi)^p =\sum _{k=0}^p \left(
    \begin{array}{c}
    p \\
    k
    \end{array}
    \right) a^{p-k} (\text{bi})^k[/itex]

    [itex]Re[(a+bi)^p] =\sum _{k=0}^p \left(
    \begin{array}{c}
    p \\
    k
    \end{array}
    \right) a^{p-k} (\text{bi})^k [/itex]

    [itex]Re[\displaystyle \sum _{k=0}^p \text{bi}^k a^{p-k} \left(
    \begin{array}{c}
    p \\
    k
    \end{array}
    \right)] = \sum _{k=0}^{p/2} \left(
    \begin{array}{c}
    p \\
    2k
    \end{array}
    \right) a^{p-2k} (\text{bi})^{2k}[/itex]

    I just thought that for each even power of bi that that part will be real. The answer is completely different though. Just confused.

    http://www.exampleproblems.com/wiki/index.php/CV8
     
    Last edited: May 25, 2009
  2. jcsd
  3. May 25, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Re: Re[I9 (a+bi)^p]

    Why don't you just convert a+bi into polar form to make it easier?
     
  4. May 25, 2009 #3
    Oh right yeah, that makes it very easy to find. In the solution the modulus isn't included though? I thought it would be that multiplied by the modulus of the a+bi bit
     
  5. May 25, 2009 #4
    For what it's worth, I think that modulus^p should be out in front of that cosine in the link. Unless the modulus is specified to be of 1 someplace.
     
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