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Math Help - complex numbers

  1. #1
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    complex numbers

    Hi is my reasoning correct?


    I want z^2=conj(Z) in the form z=a+bi.

    What I think is z=sqrt[conj(Z)].


    Then


    Z= sqrt[a-bi].


    Z=sqrt(a) – sqrt(b)i.

    Thanks………………
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  2. #2
    A Plied Mathematician
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    I would multiply out the LHS thus:

    (a+bi)(a+bi)= a^2 + 2abi - b^2, and equate real and imaginary components to the RHS: a-bi. Two equations and two unknowns.
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  3. #3
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    response

    So if



    (a+bi)(a+bi)= a^2 + 2abi - b^2


    Then


    Z= a^2 + 2abi or a(a +2bi)



    or


    Z= -(b)^2 + 2abi or -b(b -2ai)
    Last edited by Neverquit; June 9th 2011 at 02:40 AM. Reason: error
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  4. #4
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    Re your first post: in general \sqrt{a+ b}\ne \sqrt{a}+ \sqrt{b}. Also are "z" and "Z" the same? It is bad practice to use capital and non-capital letters interchangeably.

    I have no idea where you got "Z= a^2+ 2abi". You seem to have just dropped the "-b^2" from the previous line. And then just dropped "a^2" to get that last line. You can't do that!

    If z= a+ bi, then z^2= a^2- b^2+ 2abi. conj(z)= a- bi so your equation, z^2= conj(z), is a^2- b^2+ 2abi= a- bi. Equating real and imaginary parts gives a^2- b^2= a and 2ab= -b.

    I get four complex number solutions.
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