Question

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))

Answer 1

Answer with explanation:

`Z=(-2 i)^{(1)/(3)}\\\\Z^3= -2 i\\\\Z^3=2[0-  i]\\\\z^3=2[\cos((3\pi )/(2))+i \sin((3\pi )/(2))]\\\\z=2^{(1)/(3)}[\cos((3\pi )/(2))+i \sin((3\pi )/(2))]^{(1)/(3)}\\\\z=2^{(1)/(3)}[\cos(2k\pi +(3\pi )/(2))+i \sin(2k\pi +(3\pi )/(2))]^{(1)/(3)}\\\\z=2^{(1)/(3)}[\cos((2k\pi)/(3) +(\pi )/(2))+i \sin((2k\pi)/(3) +(\pi )/(2))]^{(1)/(3)}\\\\ \text{use Demoiver's theorem}}\\\\ (\cos A + \sin A)^n=\cos nA +i\sin nA\\\\ \text{for, k=0,}}`

we will get one root of

`(-2 i)^{(1)/(3)},\text{which is equal to }}=2^{(1)/(3)}[\cos((\pi )/(2))+i \sin(\pi )/(2))]`

Option 3

Answer 2

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.