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# Complex Number Problem

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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.

Feb 4, 2024

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

Feb 6, 2024