Complex numbers as roots of quadratic equation

Question

I'm trying to solve quadratic equation

$$l^{2} - (2\cos x)l + 1=0$$

so x's are $$ l=\frac{2\cos x\pm\sqrt{4\cos^{2}x-4}}2 $$ I've checked the solution on online calculators and they say it's $\cos x \pm i\sin x$. So my question is how can I solve equation above to get these roots?

Answer 1

Continuing with what you have: $$ \frac{2\cos x\pm\sqrt{4\cos^{2}x-4}}2 = \frac{2\cos x\pm\sqrt{-4\sin^{2} x}}2 \\= \frac{2 \cos x \pm 2i \sin(x)}{2} = \cos x \pm i \sin x \, . $$

Or, without that detour: $$ l ^2 -(2\cos x) l+1 = l ^2 -(2\cos x) l+ \cos^2 x + \sin^2 x \\ = (l-\cos x)^2 - (i\sin x)^2 $$ is zero if and only if $$ l - \cos x = \pm i \sin x \iff l = \cos x \pm i \sin x $$

Answer 2

The roots can be found with the quadratic formula. Note for your quadratic, we have $a = 1, b= -2cos(x), c = 1$. This gives us the roots via the quadratic formula: \begin{align} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} &= \frac{2cos(x) \pm \sqrt{4cos^2(x) - 4}}{2}\\ &= \frac{2cos(x) \pm 2\sqrt{cos^2(x) - 1}}{2}\\ &= cos(x) \pm \sqrt{cos^2(x) - 1}\\ &= cos(x) \pm \sqrt{-1(1 - cos^2(x))}\\ &= cos(x) \pm i\sqrt{1 - cos^2(x)}\\ &= cos(x) \pm i\sqrt{sin^2(x)}\\ &= cos(x) \pm i sin(x)\\ \end{align}

Answer 3

$$l^{2} - (2\cos x)l + 1=0 $$

$$ l=\frac{2\cos x\pm\sqrt{4\cos^{2}x-4}}2 $$

The discriminant is negative. So, simplifying ( The common factor 2 cancels in the numerator and denominator ) and the above $ \sqrt{-1}=i$ should occur in the complex roots, and as $ \cos^2 x + \sin^2 x =1,$ we have

$$ l= \cos x + i \sin x $$

The roots are in the complex plane, the parabola axis may be plotted parallel to real x-axis with imaginary y-axis thus: