How to solve a quadratic that needs complex numbers?

Question

I've been given a quadratic equation that I know will have complex roots but I can't figure out how to get there.

The equation is

$$x^2 - 2x\cos(\alpha) + 1$$

I put it into the quadratic formula and got

$$\frac{\cos(\alpha) \pm \sqrt{\cos^2(\alpha) - 1}}{x}$$

but I don't know where to go from here.

We've been told the answer is $$\cos(\alpha) \pm i\sin(\alpha)$$ but I don't know why!

Answer 1

Rewrite as $$\frac{\cos(\alpha) \pm \sqrt{-\sin^2 \alpha} }{\color{red}{1}}$$ $$=\\cos(\alpha) \pm i\sin \alpha$$

Note:

1) $\sin^2x =1-\cos^2x$

2) Correction in quadratic formula

Answer 2

Hint. Recall that $\cos^2\alpha=1-\sin^2\alpha,$ so that $$\cos^2\alpha-1=-\sin^2\alpha.$$ Hopefully you can take it from here.