Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.

Question

Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.

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Kim Chi 2 months 2021-07-22T21:53:12+00:00 1 Answers 6 views 0

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    2021-07-22T21:54:45+00:00

    Answer:

    z^{\frac{1}{3} }= -0.978 + i\cdot 2.836, z^{\frac{1}{3} }= -1.967 - i\cdot 2.265, z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

    Explanation:

    The cube root of the complex number can determined by the following De Moivre’s Formula:

    z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]

    Where angles are measured in radians and k represents an integer between 0 and n - 1.

    The magnitude of the complex number is 27 and the equivalent angular value is 1.817\pi. The set of cubic roots are, respectively:

    k = 0

    z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]

    z^{\frac{1}{3} }= -0.978 + i\cdot 2.836

    k = 1

    z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]

    z^{\frac{1}{3} }= -1.967 - i\cdot 2.265

    k = 2

    z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]

    z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

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