If $(\sqrt{3}+i)^{100}=2^{99}(p+i q)$, then $p$ and $q$ are roots of the equation:

(D) $x^{2}-(\sqrt{3}-1) x-\sqrt{3}=0$

(A) $x^{2}+(\sqrt{3}+1) x+\sqrt{3}=0$

(B) $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$

(C) $x^{2}+(\sqrt{3}-1) x-\sqrt{3}=0$

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Complex Numbers

(3 + i)^{100} = 2^{99} (p + i q)

, then

p

and

q

are roots of the equation:

Answer: (D)

Writing

3 + i

in polar form,

3 + i = 2 (cos \frac{π}{6} + i sin \frac{π}{6})

, so by De Moivre's theorem

(3 + i)^{100} = 2^{100} (cos \frac{100 π}{6} + i sin \frac{100 π}{6})

Since

\frac{50 π}{3} = 16 π + \frac{2 π}{3}

, this equals

2^{100} (cos \frac{2 π}{3} + i sin \frac{2 π}{3}) = 2^{100} (- \frac{1}{2} + i \frac{3}{2}) = 2^{99} (- 1 + i 3)

Comparing with

2^{99} (p + i q)

gives

p = - 1

and

q = 3

, whose sum is

3 - 1

and whose product is

- 3

. The quadratic with these two numbers as roots is

x^{2} - (3 - 1) x - 3 = 0

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option