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))

Question

Johnny

Mathematics

5 January, 05:04

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))

+3

Answers (1)

Know the Answer?

Cale Fitzpatrick · Answer 1

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.

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))

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))