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How to find the complex cube roots of -1?

  • 2 years ago
  • 2 years ago

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  1. vengeance921
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    -1=i^2 so complex number cube root would be \[\sqrt[3]{i ^{2}}\] \[(i ^{2})^{1/3}\] \[i ^{2/3}\] \[i ^{2}-i ^{3}\] \[-1-\sqrt{-1}\] but the final answer would just be until i^2/3 i think since it is asked in terms of complex numbers

    • 2 years ago
  2. rivermaker
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    It may be simpler to do as follows: \[x^3= -1 \rightarrow x^3+1=0 \rightarrow (x+1)(x^2-x+1)=0\] This gives x = -1 as the real root and the two complex roots can be obtained by solving the quadratic equation\[x^2-x+1\]

    • 2 years ago
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