// Numbas version: exam_results_page_options {"name": "De Moivre's theorem 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "

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]]

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