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!

  • 1
    $\begingroup$ Two things: 1) you're missing the "fundamental trig identity" $sin^2(a) + cos^2(a) = 1$. (2) you don't need the $x$ in the denominator; you need its coefficient, which is $1$ $\endgroup$ May 6, 2020 at 20:57
  • $\begingroup$ You misapplied the formula. You shouldn't have the $x$ in the denominator. It's only a $1.$ $\endgroup$
    – Allawonder
    May 6, 2020 at 20:58

2 Answers 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.


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


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

2) Correction in quadratic formula


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