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Quick complex number question

  1. Oct 14, 2009 #1
    4z + z(bar) = 5 + 9i
    then z = [solve for this]


    I don't know how to re arrange this equation

    (5+9i)/4 = z+z(bar)
     
  2. jcsd
  3. Oct 14, 2009 #2

    lurflurf

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    z+z(bar)=2Re(z)
    but written
    4z + z(bar) = 5 + 9i
    is not the same as
    (5+9i)/4 = z+z(bar)
    but
    4(z + z(bar) )= 5 + 9i
    is
     
  4. Oct 14, 2009 #3

    Dick

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    Just write z=a+bi and separate it into real and imaginary parts.
     
  5. Oct 14, 2009 #4

    Mentallic

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    lurflurf, I can't see where you are headed with your approach.

    [tex]4(z+\bar{z})=4z+4\bar{z} \neq 4z+\bar{z}[/tex]

    vorcil take Dick's approach. Just remember that [itex]\bar{z}[/itex] is the complex conjugate of z, thus if z=a+ib then [itex]\bar{z}[/itex]=a-ib
     
  6. Oct 15, 2009 #5
    4z + z(bar) = 5 + 9i

    4z(re) + z(bar)(re) = 5
    4z + zbar = 5, z(re) = 1 since zbar(re) = z(re)

    4z(im) + z(bar)(im) = 9i
    4x - x = 9
    z(im) = 3i

    z = 1 + 3i
    checking
    4(1+3i) + (1-3i) = 5+9i

    is that right?
     
  7. Oct 15, 2009 #6

    Mentallic

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    Yes that is correct, except just remember to be sure to re-read the question so you answer exactly what it asked. "solve for z"
    So then just at the end write: z=1+3i

    Also, just another similar method which you might find simpler and this usually makes a larger variety of questions more simple and easy to understand:

    [tex]4z+\bar{z}=5+9i[/tex]

    let z=a+ib

    [tex]4a+4ib+a-ib=5+9i[/tex]

    [tex]5a+3ib\equiv 5+9i[/tex]

    Hence,
    [tex]5a=5, a=1[/tex]
    [tex]3ib=9i, b=3[/tex]

    Therefore, [tex]z=1+3i[/tex]
     
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