// Numbas version: exam_results_page_options {"name": "De Moivre's theorem - Square Root", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Calculating the square root of a complex number using De Moivre.

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Use De Moivre's theorem to find a square root of \\(Z=\\var{x}+\\var{y}i\\)

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\\(\\sqrt{\\var{x}+\\var{y}i}=\\pm(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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To evaluate \\(\\sqrt{\\var{x}+\\var{y}i}\\) we must first express \\(Z=\\var{x}+\\var{y}i\\)  in polar form.

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The modulus of \\(\\var{x}+\\var{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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