+0 The polynomial P(x) is a monic, quartic polynomial with real coefficients, and two of its roots are
cosθ+isinθ and sinθ+icosθ where 0<θ<π4. 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.
+410 cos(θ)+isin(θ)=eiθ by Euler's formula, or representing the point on the unit circle with terminal side with degree θ.
Similarly,
sin(θ)+icos(θ)=cos(π/2−θ)+isin(π/2−θ)=ei(π/2−θ), and this represents the angle with terminal side with degree π/2−θ.
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−iθand ei(θ−π/2)=ei(3π/2+θ)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(θ)+2cos(θ))∗(cos(θ)−sin(θ))/2=cos2(θ)−sin2(θ)=cos(2θ)
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, eiθ∗e−iθ∗ei(π/2−θ)∗ei(θ−π/2)=e0=1
So P(0) is equal to 1
The Area is half of P(0) which is 1/2 so cos(2θ)=1/2
2θ=π/3
θ=π/6
The sum of the roots is, 2cos(π/6)+2sin(π/6)=√3+1
