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Homework Help: Complex function conjugate

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    I have a complex function


    What is the conjugate?

    2. The attempt at a solution

    The conjugate is


    My question is, is my answer correct? If I wrong, could you tell me what is the right answer and how to reach this correct answer?
  2. jcsd
  3. Jan 26, 2010 #2
    Your answer is flawless!

  4. Jan 26, 2010 #3
    Thanks. I was sure about that, but one of my friend said I was wrong. So, is there a way to proof it? Because I see it as a definition that whenever there is a complex function [tex]w\left(z\right)=w\left(x+iy}\right)[/tex], then the conjugate is always [tex]w\left(z^{*}\right)=w\left(x-iy\right)[/tex]. I would be more than happy if someone can show me or point me to a proof that the above definition is always true.
  5. Jan 26, 2010 #4
    But I think you need
    and not

    Ask your advisor B. .....L.P. !!
    Last edited: Jan 26, 2010
  6. Jan 26, 2010 #5
    Since that is a definition, we can't do so much to give a real proof. But you can imagine the case involving [tex]e^{i\theta}[/tex]. In this case, using the Euler relation one can see

    [tex]e^{i(-\theta)}=\cos(\theta)-i\sin(\theta)[/tex] (by the symmetry of cos and anti-symmetry of sin wrt the change of the sign of [tex]\theta[/tex]),


    [tex]e^{i\theta}e^{i(-\theta)}=[\cos(\theta)+i\sin(\theta)][\cos(\theta)-i\sin(\theta)] = 1.[/tex].

    Here you can understand that the result of the product of the function [tex]e^{i\theta}[/tex] and its complex conjugate leads to the correct answer 1 just by looking at the first part of the equality. The second part confirms the complex conjugation of [tex]e^{i\theta}[/tex] and [tex]e^{-i\theta}[/tex]. Thus a change in the sign of [tex]i[/tex] makes the function be transferred into its complex conjugate phase.

    Last edited: Jan 26, 2010
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