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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.
\nThe modulus of \\(\\var{x}+\\var{y}j\\) = \\(\\sqrt{\\var{x}^2+(\\var{y})^2}=\\sqrt{\\simplify{{x^2}+{y}^2}}\\)
\nThe argument of the complex number is given by \\(\\theta=tan^{-1}\\left(\\frac{\\var{y}}{\\var{x}}\\right)=\\var{theta}\\)
\nAccording 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)\\)
\nBy letting \\(\\theta\\) take the values 0, 1 and 2 we find each of the three cube roots.
\nBelow is the third cube root which can be found by setting \\(\\theta=1\\).
\n\\((\\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)\\)
\n=\\(\\simplify{{x4}+{y4}j}\\)
\n\\(A=\\var{x4}\\) and \\(B=\\var{y4}\\)
\n", "parts": [{"prompt": "\\(Z=\\var{x}+\\var{y}j\\) has three cube roots.
\n\\(Z=\\var{x2}+\\var{y2}j\\) and \\(Z=\\var{x3}+\\var{y3}j\\) are two of the roots.
\nThe third cube root is \\(Z=A+Bj\\)
\nCalculate \\(A\\) and \\(B\\)
\n\n\\(A\\) = [[0]]
\n\\(B\\) = [[1]]
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