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Exponential Form Question

  1. Sep 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that

    [tex]\overline{e^{i\theta}} = e^{-i\theta}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    So I what's going through my mind is that the problem above is pretty much the same as saying [tex]\bar{z} = z^{-1}[/tex]

    Then to prove it is all I need to say is that since [tex]\overline{e^{i\theta}} = (cos\theta - isin\theta)[/tex] and [tex]e^{-i\theta} = (cos\theta - isin\theta)[/tex]

    so then they are equal. Is this sufficient or am I totally under thinking it?
  2. jcsd
  3. Sep 2, 2009 #2
    equal is equal. That's a silly problem. maybe show that bar above [tex]cos \theta + i sin \theta[/tex] just to be on the safe side.
  4. Sep 2, 2009 #3


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    Homework Helper

    Well that is how I would do it.
  5. Sep 2, 2009 #4
    ignore my last post, do [tex]e^{-i\theta} = cos( -\theta) + i sin (-\theta) [/tex] and take it from there.

    The reason being that you want to apply any factors in the exponent to [tex]\theta[/tex] rather than to i.
  6. Sep 2, 2009 #5
    my opinion is to use trigonometry to prove exp functions and try to use reverse; i.e exp functions to prove trigonometry in complex analysis. thats my suggestions.
    esp. Euler formula
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