Let p,q∈ℝ and ( 1 - √3 i )²⁰⁰ = 2¹⁹⁹(p + iq),i = √(- 1) then p+q+q² and p-q+q² are roots of the equation.

Solution

Maths sir · 4 steps

Put 1-√3i in polar form first, and De Moivre turns a 200th power into one simple angle.

1. Polar form. The modulus is √(1+3) = 2, and the point (1, -√3) lies in the fourth quadrant at angle -π/3. So 1-√3i = 2(cos π/3 - i sin π/3).

   (1-√3i)²⁰⁰ = 2²⁰⁰(cos 200π/3 - i sin 200π/3)

2. Reduce the angle. 200π/3 = 66π + 2π/3, and 66π is a whole number of full turns. So cos 200π/3 = -1/2 and sin 200π/3 = √3/2, which gives 2²⁰⁰(-1/2 - √3/2i) = 2¹⁹⁹(-1 - √3i).
3. Find the roots. Comparing, p = -1 and q = -√3, so q² = 3. Then p+q+q² = 2-√3 and p-q+q² = 2+√3.
4. So. Sum of roots = 4 and product = (2-√3)(2+√3) = 4-3 = 1. The equation is x² - 4x + 1 = 0, which is option (b).

The idea to keep. For a big power of a+bi, convert to r(cos θ + i sin θ), multiply the angle, then strip off full turns of 2π. Build the quadratic as x² - (sum)x + product.

Answer from the NTA final answer key. Solution by PrepWiser.