Question

Which expression is a cube root of -1+i sqrt3

Answer 1

If
`z=-1+i √(3)`, then


`Rez=-1`

`Imz= √(3)` and

`|z|= √(Re^2z+Im^2z)= \sqrt{1^2+ (√(3))^2 }= √(4)=2`.
Then
`cos\theta = (Rez)/(|z|) = (-1)/(2)`
`sin\theta = (Imz)/(|z|)= ( √(3) )/(2)`.

Since
`\theta \in (-\pi,\pi]`, you can conclude that
`\theta= (2\pi)/(3)`.
The triginometric form of z is
`z=|z|(cos\theta+isin\theta)=2(cos (2\pi)/(3) +isin (2\pi)/(3) )`.
Use formula
`\sqrt[3]{z} =\{ \sqrt[3]z (cos (\theta+2\pi k)/(n)+isin (\theta+2\pi k)/(n) ), k=0,1,2\}` to find cube root.
For k=0,
`z_1= \sqrt[3]{2} (cos ( (2\pi)/(3) )/(3) +isin( (2\pi)/(3) )/(3))= \sqrt[3]{2} (cos 40^0+isin 40^0)`,
For k=1,
`z_1= \sqrt[3]{2} (cos ( (2\pi)/(3)+2\pi )/(3) +isin( (2\pi)/(3)+2\pi )/(3))= \sqrt[3]{2} (cos 160^0+isin 160^0)`,
For k=2,
`z_1= \sqrt[3]{2} (cos ( (2\pi)/(3)+4\pi )/(3) +isin( (2\pi)/(3)+4\pi )/(3))= \sqrt[3]{2} (cos 280^0+isin 280^0)`.
Answer: The correct choice is C.