Complex roots simple question -quick! :)

For some reason I'm getting roots which aren't complex conjugates of each other, so I think I've messed up with the angles.

So I'm pretty sure I've done a wrong step just before the roots solution. Can someone please guide me on the path back to the correct answer?

Hi everyone,

I'm trying to figure out the solutions to $z^4 = -2 + 2\sqrt3i$

For some reason I'm getting roots which aren't complex conjugates of each other, so I think I've messed up with the angles.

In polar form: r=4, $\theta=2\pi/3$

Therefore $cos4\theta=cos2\pi/3$

So the solutions are $\theta= \pi/6 + (k\pi/2)$ for k=0,1,2,3

And then after that my solutions aren't conjugates i.e
z1 = $\sqrt2(cos(\pi/6) + isin(\pi/6))$
And then add pi/2 to each angle until the last solution:
z4 = $\sqrt2(cos(5\pi/3)+isin(5\pi/3)$

So I'm pretty sure I've done a wrong step just before the roots solution. Can someone please guide me on the path back to the correct answer?

The roots of a polynomial are only necessarily complex conjugates of each other if the coefficients of the polynomial are real.

As joostan said but also, if you are allowed a calculator such as the Casio 991es then you can put it in complex mode, enter your z and type AnsAnsAnsAns to confirm that you were correct.