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Use De Moivre's theorem to calculate \\((\\simplify{+{x}+{y}i})^\\var{n}\\)

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Calculating complex numbers raised to an natural number exponent

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To evaluate \\((\\simplify{{x}+{y}i})^\\var{n}\\) we must first express \\((\\simplify{{x}+{y}i})\\) in polar form.

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The modulus of \\(\\simplify{+{x}+{y}i}\\) = \\(\\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^{\\var{n}}=|Z|^{\\var{n}}\\left(\\cos(\\var{n}*\\theta)+i\\sin(\\var{n}*\\theta)\\right)\\)

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\\((\\simplify{{x}+{y}i})^\\var{n}=\\left(\\sqrt{\\simplify{{x^2}+{y}^2}}\\right)^{\\var{n}}\\left(\\cos(\\simplify{{n}*{theta}})+i\\sin(\\simplify{{n}*{theta}})\\right)\\)

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=\\(\\simplify{{x2}+{y2}i}\\)

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\\(A=\\var{x2}\\)  and  \\(B=\\var{y2}\\)

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Note that the real and imaginary parts of your answer should be integers (whole numbers). 

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\\((\\simplify{{x}+{y}i})^\\var{n}=A+Bi\\)

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

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\\(A\\) = [[0]]

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\\(B\\) = [[1]]

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