The polynomial P(x) is a monic, quartic polynomial with real coefficients, and two of its roots are

\(\cos \theta + i \sin \theta\) and \( \sin \theta + i \cos \theta\) where \(0 < \theta < \frac{\pi}{4}\). When the four roots of P(x) are plotted in the complex plane, they form a quadrilateral whose area is equal to half of P(0). Find the sum of the four roots.

MathIAExploration Feb 4, 2024

#1**+3 **

\(cos( \theta)+isin(\theta) = {e}^{i\theta}\) by Euler's formula, or representing the point on the unit circle with terminal side with degree \(\theta\).

Similarly,

\(sin(\theta)+icos(\theta) = cos(\pi/2 - \theta) + isin(\pi/2 - \theta) = {e}^{i(\pi/2 - \theta)}\), and this represents the angle with terminal side with degree \(\pi/2 - \theta\).

If a complex number is a root of a polynomial, we also know that the conjugate is a root of a polynomial.

Therefore, we know that, \({e}^{-i\theta}\)and \({e}^{i(\theta-\pi/2)} = {e}^{i(3\pi/2 + \theta)}\)are also roots.

We see that these roots form a trapezoid. (stuff up there not completely necessary. it was my first instinct).

The formula for the area of a trapezoid, is (top + bottom)*height/2.

Finding these values, and plugging in, \(Area = (2sin(\theta)+2cos(\theta))*(cos(\theta) - sin(\theta))/2 = {cos}^{2}(\theta)-{sin}^{2}(\theta)=cos(2\theta)\)

First we know that P(0) is equal to the last term, or the constant term, which is also equal to the product of the roots by Vietas,

Multiplying, \({e}^{i\theta}*{e}^{-i\theta}*{e}^{i(\pi/2-\theta)}*{e}^{i(\theta-\pi/2)} = {e}^{0} = 1\)

So P(0) is equal to 1

The Area is half of P(0) which is 1/2 so \(cos(2\theta)=1/2\)

\(2\theta = \pi/3\)

\(\theta=\pi/6\)

The sum of the roots is, \(2cos(\pi/6)+2sin(\pi/6)=\sqrt{3}+1\)

hairyberry Feb 6, 2024