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Math Help - Roots of an equation

  1. #1
    MHF Contributor arbolis's Avatar
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    Roots of an equation

    I must find the root of z^3+1=0.
    I both used the polar and Cartesian approaches and reached the result that the root is z=-1.
    Is it possible? Shouldn't there be 3 roots? (they could be multiple roots I guess).
    I can show the details of my attempt if you like.
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  2. #2
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    You have found only one root !
    Now you can factor z^3+1 = (z+1)(z^2-z+1) and solve z^2-z+1=0 to find the two other roots
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  3. #3
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    Anyway you should have found 3 roots

    z^3 = -1

    r^3 \:e^{3i \theta} = e^{i \pi}

    r^3=1 and 3 \theta = \pi + 2k \pi

    r=1 and \theta = \frac{\pi}{3} + \frac{2k \pi}{3}

    z_1 = e^{i \frac{\pi}{3}}

    z_2 = e^{i \pi} = -1

    z_3 = e^{i \frac{5 \pi}{3}} = e^{-i \frac{\pi}{3}}
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