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Quick question about conjugates

  1. Sep 21, 2010 #1
    Hi I have a parametrisation of the unit circle
    z = cos(t) + isin(t) = e^(it)
    now i guess that the conjugate of that is the same but with a negative coefficient on the imaginary part, namely:
    conj z = cos(t) - isin(t)
    How does that extend to the e^(it) form? is the following true:
    cos(t) - isin(t) = e^(-it)

    Thanks
     
    Last edited: Sep 21, 2010
  2. jcsd
  3. Sep 21, 2010 #2

    Office_Shredder

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    [tex]e^{-it}=\cos(-t)+i\sin(-t)[/tex]. cosine is an even function, an sine is an odd function, so

    [tex]e^{-it}=\cos(t)-i\sin(t)[/tex]
     
  4. Sep 23, 2010 #3
    nice, thanks.
     
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