If ( √3 + i )¹⁰⁰ = 2⁹⁹(p + iq), then p and q are roots of the equation :

Solution

Maths sir · 4 steps

Put √3 + i in polar form first, and the 100th power becomes a single angle to reduce.

1. Polar form. The modulus is √(3 + 1) = 2 and the argument is π/6. So √3 + i = 2(cos π/6 + i sin π/6).

   (√3 + i)¹⁰⁰ = 2¹⁰⁰(cos 100π/6 + i sin 100π/6)

2. Reduce the angle. 100π/6 = 50π/3 = 16π + 2π/3, and 16π is 8 full turns. So the angle is 2π/3, with cos 2π/3 = -1/2 and sin 2π/3 = √3/2.
3. Find p and q. 2¹⁰⁰(-1/2 + √3/2i) = 2⁹⁹(-1 + √3 i), so p = -1 and q = √3.
4. So. The sum of the roots is √3 - 1 and the product is -√3. The equation is x² - (√3 - 1)x - √3 = 0, which is option (a).

The idea to keep. For a big power of a complex number, switch to polar form, multiply the angle, then strip off full 2π turns. For the quadratic, use x² - (sum)x + product = 0.

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