1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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)*
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook