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Conjugate of e^(iz)

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    I want to show that

    [tex] \exp \left( i\bar{z} \right) = \overline{\exp \left( iz \right)} [/tex]

    if and only if

    [tex] z = n\pi [/tex]

    for any integer n.

    2. Relevant equations



    3. The attempt at a solution
    Utilizing Euler's formula, I got

    [tex] \cos \bar{z} = \cos z [/tex]

    and

    [tex] \sin \bar{z} = -\sin z [/tex]

    Though not fully convinced, I concluded that

    [tex] \bar{z} = z [/tex]

    This then led me to

    [tex] \sin z = 0 [/tex]

    This obviously led me to the needed conclusion. But was I correct?
     
  2. jcsd
  3. Mar 9, 2009 #2

    Office_Shredder

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    Euler's formula is only valid for z a real number. Taking conjugates doesn't get you very far there. You should split z into x+iy and divvy up the exponential first
     
  4. Mar 9, 2009 #3
    But Euler's formula has already been proven to apply even for complex numbers. My primary concern is whether or not

    [tex]
    \exp \left( i\bar{z} \right) = \overline{\exp \left( iz \right)}
    [/tex]

    would give me

    [tex]
    \bar{z} = z
    [/tex]
     
  5. Mar 9, 2009 #4

    Office_Shredder

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    I didn't mean valid there, I meant useful.

    For example:

    [tex] \cos \bar{z} = \cos z [/tex]

    and

    [tex] \sin \bar{z} = - \sin z [/tex]

    is satisfied by any imaginary number (just by using the even/oddness of cosine/sine).

    The problem though, is that if z is a complex number, you don't have that cos(z) and sin(z) are real numbers, so you can't draw the two relations you have
     
  6. Mar 9, 2009 #5

    lurflurf

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    exp(i z)*=exp((i z)*)=exp(i*z*)=exp(-i z*)
    it is also easy to see
    exp(i z)=exp(-Im[z]+i Re[z])=exp(-Im[z])exp(i Re[z])
     
  7. Mar 9, 2009 #6

    gabbagabbahey

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    That's a strange conclusion:wink:.....If I told you [itex]\cos(0)=\cos(100\pi)[/itex], would you then conclude [itex]0=100\pi[/itex]?
     
  8. Mar 9, 2009 #7
    Alright. What if I say

    [tex]
    \cos \bar{z} = \cos z
    [/tex]

    would give me

    [tex]
    \bar{z} = z + 2n\pi
    [/tex]

    (Is this correct?)

    I will then use it to say that

    [tex]
    \sin \bar{z} = -\sin z
    [/tex]

    gives me

    [tex] \sin \left( z + 2n\pi \right) = -\sin z [/tex]

    This implies that

    [tex] \sin z = -\sin z [/tex]

    Therefore,

    [tex] z = n\pi [/tex]
     
  9. Mar 10, 2009 #8

    lurflurf

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    Why even use sine?
    exp(i z*)=exp(-i z*)
    exp(2i z*)=1
    z=n pi
     
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