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Math Help - Another complex number

  1. #1
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    Another complex number

    If z=x+yi and z^2=a+bi , x,y,a,b are all real numbers . Prove that 2x^2=\sqrt{(a^2+b^2)}+a
    By solving the equation z^4+6z^2+25=0 for z^2 or otherwise , express each root of the equation in the form of x+yi
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    If z=x+yi and z^2=a+bi , x,y,a,b are all real numbers . Prove that 2x^2=\sqrt{(a^2+b^2)}+a
    By solving the equation z^4+6z^2+25=0 for z^2 or otherwise , express each root of the equation in the form of x+yi
    Hi

    If z=x+yi then
    z^2=x^2-y^2+2xyi

    Then
    (i) x^2-y^2=a
    (ii) 2xy=b then y=\frac{b}{2x}

    Substituting y in (i)
    x^2 - \frac{b^2}{(2x)^2}=a

    (2x^2)^2 - 2a (2x^2) - b^2=0

    Reduced discriminant is a^2 + b^2

    The only positive solution is therefore 2x^2=\sqrt{a^2+b^2}+a

    Equation z^4+6z^2+25=0 for z^2
    gives z^2 = -3 + 4i or z^2 = -3 - 4i

    It is just an application with a=-3, b=4 and then a=-3 and b=-4
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