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Homework Help: Complex conjugate

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data
    i am supposed to prove that for the complex number z=cis[tex]\theta[/tex]
    the conjugate is [tex]\frac{1}{\overline{z}}[/tex]


    2. Relevant equations
    if
    z=a+bi
    [tex]\overline{z}[/tex]=a-bi

    3. The attempt at a solution
    all that i can think of is that [tex]\frac{1}{cos\theta i sin \theta}[/tex]
    =(cos [tex]\theta[/tex] i sin [tex]\theta[/tex])-1

    i have also just tried it with a random complex number such as w=2+3i
    still how does [tex]\overline{w}[/tex]=1/2+3i ????

    Im very lost.....
     
  2. jcsd
  3. Mar 14, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    [tex]\frac{1}{\overline z}[/tex]
    is NOT equal to
    [tex]\frac{1}{cos(\theta)isin(\theta)}[/tex]

    Also your problem, as stated, is wrong- the complex conjugate of [itex]z= cis(\theta)[/itex] is not 1 over the complex conjugate of z.
    [tex]\frac{1}{\overline{z}= z[/tex]
    or
    [tex]\frac{1}{z}= \overline{z}[/tex].

    The complex conjugate of [itex]cis(\theta)= cos(\theta)+ i sin(\theta)[/itex] is [itex]cos(\theta)- i sin(\theta)[/itex].

    The reciprocal of that is, of course,
    [tex]\frac{1}{cos(\theta)- i sin(\theta)}[/tex]

    Now, "rationalize the denominator"- multiply both numerator and denominator by [itex]cos(\theta)+ i sin(\theta)[/itex]

    Conversely,
    [tex]\frac{1}{z}= \frac{1}{cos(\theta)+ i sin(\theta)}[/tex]

    Multiply both numerator and denominator by [itex]cos(\theta)- i sin(\theta)[/itex].
     
    Last edited by a moderator: Mar 14, 2010
  4. Mar 14, 2010 #3
    Ah that makes it alot easyer, i totally forgot about when the denominator is imaginary that you multiply both numerator and denomiator by the conjugate.
    Thanks alot!!
     
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