Quadratic equations with complex coefficients I am stuck on the final bit of this question. There is an answer for this on stack exchange but is different from two other answers on others sites which are also different to each other. 
Solve the equation $z^2=-\sqrt3 + i$
So far i've done:
$$|z| = 2$$
$$\theta = \dfrac{5\pi}{6}$$
$$\therefore 2cis\dfrac{5\pi}{6}$$
De Moivre's theorem: $$z^2=r^2cis2\theta$$
$$\therefore$$
$$r^2=2$$
$$r = \sqrt2$$
$$cis2\theta = cis\dfrac{5\pi}{6}$$
$$2\theta = \dfrac{5\pi}{6}$$
$$\theta = \dfrac{5\pi}{12}$$
$$\therefore z = \sqrt2cis(\dfrac{5\pi}{12} + \pi n)$$
for any integer n 
If that is all correct, (please correct if it isn't), how do I finish it off? My textbook explains something about two distinct values.
I can leave it in polar form. As I said, I've seen three completely different answers to this. 

Nearly or I should say almost perfect. 
$$z^2=2e^{i(\frac {5\pi}{6}+2\pi k)}$$
Hence $$z=\sqrt 2e^{i(\frac {5\pi}{12}+\pi k)}$$
For $k=0,1$
We consider only 2 values because since we are dealing with a quadratic we must get only two roots.
In general if we find solution to some $z^n=re^{i\theta+2\pi k}$
Then $k=0,1,2,....,(n-1)$
Hence the answer is $z_1=\sqrt 2e^{\frac {5i\pi}{12}}$ and $z_2=\sqrt 2e^{\frac {17i\pi}{12}}$
