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Complex Conjugates: f*(z) = f(z*)

  1. Jun 18, 2009 #1
    It's commonly known that if f(z) is analytic, then

    f(z*) = f*(z)

    that is, an analytic function of the complex conjugate is equal to the complex conjugate of the function...with the proviso that f(x+i0) = f(x) = Re f(x)

    I've tried to prove it using the C-R equations but I'm not having much luck. Can anyone point me in the right direction?

  2. jcsd
  3. Jun 18, 2009 #2


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    If a function is analytic, then near each point in its domain, it's equal to its Taylor series (which has a positive radius of convergence)
  4. Jun 25, 2009 #3
    This is known as the Reflection Principle in textbooks.

    The two functions [tex]f(x)[/tex] and [tex]g(x) := \overline{f(\overline{x})}[/tex] are both analytic and they agree on the real line, which is a set with a limit point, therefore they agree everywhere.
  5. Jul 5, 2009 #4

    Should have spotted the method where you write it as a Taylor series (necessarily with real coefficients, I guess, since the function's restriction to the real line must be real, as stated above) then apply complex conjugation to this series - then it just boils down to showing that the complex conjugate of a complex variable raised to a power is the power of the complex congugate of the variable. In other words

    (z^n)^{*} = (z^{*})^n

    which isn't too hard, I hope!

    The Identity Theorem method is a bit more sublte, but it makes sense.

    Thanks Hurkyl, g_edgar.
  6. Jul 5, 2009 #5


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    Note that this is actually trivial if you use the polar form.
  7. Jul 5, 2009 #6
    Yep, was just thinking that.
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