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To evaluate \\((\\var{x}+\\var{y}j)^{\\frac{1}{3}}\\) we must first express \\(Z=\\var{x}+\\var{y}j\\)  in polar form.

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The modulus of \\(\\var{x}+\\var{y}j\\) = \\(\\sqrt{\\var{x}^2+(\\var{y})^2}=\\sqrt{\\simplify{{x^2}+{y}^2}}\\)

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The argument of the complex number is given by \\(\\theta=tan^{-1}\\left(\\frac{\\var{y}}{\\var{x}}\\right)=\\var{theta}\\)

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According to De Moivre's theorem \\(Z^{\\frac{1}{3}}=|Z|^{\\frac{1}{3}}\\left(cos(\\frac{\\theta+2n\\pi}{3})+jsin(\\frac{\\theta+2n\\pi}{3})\\right)\\)

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By letting \\(\\theta\\) take the values 0, 1 and 2 we find each of the three cube roots.

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Below is the third cube root which can be found by setting \\(\\theta=1\\).

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\\((\\simplify{{x}+{y}j})^{\\frac{1}{3}}=\\left(\\sqrt{\\simplify{{x^2}+{y}^2}}\\right)^{\\frac{1}{3}}\\left(cos(\\simplify{{n}*({theta}+2*pi)})+jsin(\\simplify{{n}*({theta}+2pi)})\\right)\\)

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=\\(\\simplify{{x4}+{y4}j}\\)

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\\(A=\\var{x4}\\)  and  \\(B=\\var{y4}\\)

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", "parts": [{"prompt": "

\\(Z=\\var{x}+\\var{y}j\\) has three cube roots.

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\\(Z=\\var{x2}+\\var{y2}j\\)   and   \\(Z=\\var{x3}+\\var{y3}j\\) are two of the roots.

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The third cube root is \\(Z=A+Bj\\)

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Calculate \\(A\\) and \\(B\\)

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\\(A\\) = [[0]]

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\\(B\\) = [[1]]

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This question asks you to use De Moivre's theorem to find the cube roots of \\(Z=\\var{x}+\\var{y}j\\)

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Calculate the cube roots of a complex number using De Moivre.

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