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Math Help - Precalc- complex numbers

  1. #1
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    Precalc- complex numbers

    The system of equations
    has two solutions
    and in complex numbers. Find .

    So far I replaced z with a+bi, did some algebra (squared both equations) and ended up with these equations:

    a^2 + b^2 + (4-4a)i + 4b = 23

    and

    a^2 + b^2 - 16a + 10b + (80 - 10a - 16b )i = -1


    Not sure where to go from here. Thanks!
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  2. #2
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    Re: Precalc- complex numbers

    This may help.
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  3. #3
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    Re: Precalc- complex numbers

    Quote Originally Posted by geisha View Post
    The system of equations
    has two solutions
    and in complex numbers. Find .

    So far I replaced z with a+bi, did some algebra (squared both equations) and ended up with these equations:

    a^2 + b^2 + (4-4a)i + 4b = 23

    and

    a^2 + b^2 - 16a + 10b + (80 - 10a - 16b )i = -1


    Not sure where to go from here. Thanks!

    You don't really want to start by squaring. The | |'s in this case stand for the magnitude of the complex expression inside them. This isn't the same as the square of that expression.

    \left \mid a + i b \right \mid = \sqrt{a^2 + b^2}

    so for example in your first equation

    \left \mid z-2-2i \right \mid = \sqrt{23}

    let z=a+bi

    \left \mid (a-2)+(b-2)i \right \mid=\sqrt{(a-2)^2+(b-2)^2}=\sqrt{23}

    do the same thing for your second equation and then you can square both sides and solve the quadratic system for a and b.
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  4. #4
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    Re: Precalc- complex numbers

    Got it, thanks!!
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