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  1. #1
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    6 distinct roots of complex number

    I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to procced?
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  2. #2
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    Re: 6 distinct roots of complex number

    Quote Originally Posted by timeforchg View Post
    I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to procced?
    Try converting to polars...

    Edit: We answered this exact question here.
    Last edited by Prove It; September 13th 2012 at 07:05 PM.
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  3. #3
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    Lightbulb Re: 6 distinct roots of complex number

    Quote Originally Posted by timeforchg View Post
    I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to procced?
    {z6}=1+\left(\frac{64 i (1-i)}{1+2 i}\right)^6

    z6 = ((64 i (1 - i))/(1 + 2 i))^6 + 1

    {zz}({k}){=}\sqrt[6]{|{z6}|} e^{i \arg \left(\frac{2 \pi  k}{6}+\frac{\text{z6}}{6}\right)}

    k = {0, 1, 2, 3, 4, 5}

    z1, z2, z3, z4, z5, z6=
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{7146825580544}{2687695088383}\right)  }
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-31250 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-62500 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{21440476741632}{8063085265149-31250 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-125000 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-156250 \pi }\right)}
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  4. #4
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    Re: 6 distinct roots of complex number

    Quote Originally Posted by MaxJasper View Post
    {z6}=1+\left(\frac{64 i (1-i)}{1+2 i}\right)^6

    z6 = ((64 i (1 - i))/(1 + 2 i))^6 + 1

    {zz}({k}){=}\sqrt[6]{|{z6}|} e^{i \arg \left(\frac{2 \pi  k}{6}+\frac{\text{z6}}{6}\right)}

    k = {0, 1, 2, 3, 4, 5}

    z1, z2, z3, z4, z5, z6=
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{7146825580544}{2687695088383}\right)  }
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-31250 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-62500 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{21440476741632}{8063085265149-31250 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-125000 \pi }\right)}
    -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-156250 \pi }\right)}

    Hi Max,

    This makes me more confused. Am suffering from brain damage! haha
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