+0 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.
+399 \(cos( \theta)+isin(\theta) = {e}^{i\theta}\) 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, \(2cos(\pi/6)+2sin(\pi/6)=\sqrt{3}+1\)
