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Complex analysis

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that (z̄ )^k =(z̄ ^k) for every integer k (provided z≠0 when k is negative)

    2. Relevant equations

    3. The attempt at a solution
    I let z=a+bi so, z̄ =a-bi
    Then I plugged that into one side of the equation to get
    I was going to try to manipulate this to get [(a-bi)^k]
    But I don't know where to go from here, or even what to do.....
    Please HELP!
  2. jcsd
  3. Jan 24, 2009 #2


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    This is a bit unclear. From the rest of your post, it looks as if by ̄ you meant the complex conjugation. So did you mean to write

    [tex] (z^*)^k = (z^{* k}) [/tex] ??

    But even this is unclear since most people would interpret the two sides to mean the same thing.

    It would make more sense if the question was to prove

    [tex] (z^*)^k = (z^k)^* [/tex]

    was that the question?

    If so, then write the complex number z in polar form [itex] r e ^{i \theta} [/itex]
  4. Jan 24, 2009 #3
    Is the bar over the z only in both parts? Or is it

    [tex] \bar{z}^{k} = \bar{\left(z^k\right)} [/tex]
  5. Jan 24, 2009 #4

    Yes, that is my question. Sorry for the lack of clarity.
    So I rewrote z in polar form, but I wasn't sure about what the conjugate of that would be...would the exponent just be negated? (the iѲ term).
    If that was the case, would I simply combine the exponents in the left side of the equation, multiplying the k and the iѲ term thus resulting in the right side of the equation?
    I hope I'm being more clear!
  6. Jan 24, 2009 #5


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    No problem!
    Yes. Complex conjugation just means that we replace any "i" we see by -i. So yes, the sign of the exponent changes.
    Exactly! Basically, k times (i)* gives the same thing as (k times i)*
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