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Math Help - Complex problem Z=z^2

  1. #1
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    Complex problem Z=z^2

    For what values a and b is the complex number z = a + bi a solution to z=z^2 ?

    I tried this in many different ways and il post one of my solutions but i dont know if its correct.
    (a+bi) = (a+bi)^2
    dividing by a+bi on both sides.
    1 = a + bi
    This tells me that Z is real and that a = 1 and b = 0.
    Is this correct?
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  2. #2
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    No, if \displaystyle z = z^2

    \displaystyle a + b\,i = (a + b\,i)^2

    \displaystyle a + b\,i = a^2 + 2ab\,i + b\,i^2

    \displaystyle a + b\,i = a^2 - b^2 + 2ab\,i

    \displaystyle 0 + 0i = a^2 - a - b^2 + (2ab - b)i.


    Therefore \displaystyle a^2 - a - b^2 = 0 and \displaystyle 2ab - b = 0.

    Solve these equations simultaneously.
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  3. #3
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    Ok.
    I recive that a=0.5 and b = +-0.25i
    Is this correct?
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  4. #4
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    \displaystyle a and \displaystyle b are both real numbers...
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  5. #5
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    Ok.
    Can you point out where im doing wrong in my calculations?
    2ab=b
    This leads me to that a = 1/2
    a^2 - a -b^2 = 0
    Here i add that a=0.5 and recive
    -0.25 = b^2
    hence i receive that b = +-0.5i
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  6. #6
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    Quote Originally Posted by Zamzen View Post
    Ok.
    Can you point out where im doing wrong in my calculations?
    2ab=b
    This leads me to that a = 1/2
    Strictly that leads to either b= 0 or a= 1/2.

    a^2 - a -b^2 = 0

    Here i add that a=0.5 and recive
    -0.25 = b^2
    hence i receive that b = +-0.5i
    Yes, which, because b must be a real number, is impossible. It follows that a cannot be 1/2 and that we must have b= 0. Now, from a^2- a- b^2= 0, with b= 0, we have a^2- a= a(a-1)= 0 so either a= 0 or a= 1.

    In fact, we could have said, from the start, that if z\ne 0, then, dividing both sides of z^2= z by z, we get z= 1. That is, z^2= z if and only if z= 0 or z= 1, real numbers.
    Last edited by HallsofIvy; December 6th 2010 at 11:19 AM.
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