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Math Help - complex numbers equation

  1. #1
    Member Jones's Avatar
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    complex numbers equation

    Hi,

    I have (z+4i)^3 + 2(z+4i)^2 - 16 = 0

    I also know that z = -2-2i is one root, and since this is an equation with real coefficients another root is -2+2i. So i need to find the third root

    Is there an easier way of solving this, other than polynomial division?
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  2. #2
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    Quote Originally Posted by Jones View Post
    I have (z+4i)^3 + 2(z+4i)^2 - 16 = 0
    I also know that z = -2-2i is one root, and since this is an equation with real coefficients another root is -2+2i. So i need to find the third root
    Look at this.
    Attached Thumbnails Attached Thumbnails complex numbers equation-cplx5.gif  
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  3. #3
    Member Jones's Avatar
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    So just expand the brackets and replace z with a+bi ?
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