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9 October, 00:43

Which expression is a cube root of - 2i?

1) 3sqrt2 (cos (210)) + isin (210))

2) 3sqrt2 (cos (260)) + isin (260))

3) 3sqrt2 (cos (90)) + isin (90))

4) 3sqrt2 (cos (60)) + isin (60))

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  1. 9 October, 03:10
    0
    To use De Moivre's theorem, we first write - 2i is cis form: 0 - 2i has r = 2 and theta = 270.

    Then we take the cube root, which means the new result will have r^ (1/3), and the angle (theta/3). This means r = cbrt (2) and theta = 90.

    This means that the answer is (cube root of 2) (cos90 + i*sin90), choice C.

    I suspect the choices "3sqrt2" is actually a cube root of 2, not 3 multiplied by the square root of 2.
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