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Math Help - Complex number question

  1. #1
    Senior Member Stroodle's Avatar
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    Complex number question

    Hi there, I'm having trouble with the following question. As usual, I'm probably missing something basic...


    If z=1+\sqrt{3}i is a solution to the equation z^3-a=0,\ a\in\ c then the other two solutions are?

    I know one of them must be z=1-\sqrt{3}i as the coefficient of z^3 is real. But I'm not sure how to find the last solution.
    From the options (mc question) it seems that it's either z=2 or z=-2.

    Thanks for your help
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  2. #2
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    Quote Originally Posted by Stroodle View Post
    Hi there, I'm having trouble with the following question. As usual, I'm probably missing something basic...


    If z=1+\sqrt{3}i is a solution to the equation z^3-a=0,\ a\in\ c then the other two solutions are?

    I know one of them must be z=1-\sqrt{3}i as the coefficient of z^3 is real. But I'm not sure how to find the last solution.
    From the options (mc question) it seems that it's either z=2 or z=-2.

    Thanks for your help
    Complex solutions always appear as conjugates.

    So if z = 1 + \sqrt{3}i is a solution, so is \overline{z} = 1 - \sqrt{3}i.


    That means z - (1 + \sqrt{3}i) and z - (1 - \sqrt{3}i) are both factors, and so is

    [z - (1 + \sqrt{3}i)][z - (1 - \sqrt{3}i)]

    =z^2 - z(1 - \sqrt{3}i) - z(1 + \sqrt{3}i) + (1 - \sqrt{3}i)(1 + \sqrt{3}i)

     = z^2 - z + \sqrt{3}iz - z - \sqrt{3}iz + 1 + 3

     = z^2 - 2z + 4.


    Since z^2 - 2z + 4 is a factor, long divide z^3 - a by z^2 - 2z + 4 and you will have the third factor. You should note that the third root will need to be real.

    Once you have the third factor, you can find a.
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