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How do you take this complex conjugate?

  1. Jul 31, 2011 #1
    What's the complex conjugate of

    [tex]
    \frac{1}{\sqrt{1+it}}, \quad t \geq 0.
    [/tex]
     
  2. jcsd
  3. Aug 1, 2011 #2
    [tex]
    \frac{1}{\sqrt{1-it}}
    [/tex]
     
  4. Aug 1, 2011 #3

    The key is that for all complex z, [itex]z \overline{z} = |z|^2[/itex] so that [itex]\overline{z} = \frac{|z|^2}{z}[/itex]

    You can see this since [itex](a + bi)(a - bi) = a^2 + b^2[/itex]


    Now [itex] |\frac{1}{\sqrt{1+it}}|^2

    = \frac{1}{|1+it|}

    = \frac{1}{\sqrt{1 + t^2}} [/itex]

    since abs commutes with multiplication, division, and exponentiation.

    and so [itex]\overline{\frac{1}{\sqrt{1+it}}}

    = \frac{\frac{1}{\sqrt{1 + t^2}}} {\frac{1}{\sqrt{1+it}}}

    = \frac {\sqrt{1+it}} {\sqrt{1 + t^2}} [/itex]

    (edit)
    Oops of course pmsrw3 is correct, and that's the same answer as mine but a lot easier! I'd delete my response but I don't want to waste all that TeX!:smile:
     
    Last edited: Aug 1, 2011
  5. Aug 1, 2011 #4

    Hurkyl

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    (Don't forget about branch cuts! A little bit of care must be used to ensure that the function and its proposed conjugate make consistent choices of principal value)
     
  6. Aug 1, 2011 #5
    It works out OK in this case :-) I did actually think about that. If you picture the operations on the complex plane, it's pretty easy to see.
     
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