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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Mathematics question on Complex Numbers and Quadratic Equations

The value of (1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3\left(\dfrac{1+\sin\dfrac{2\pi}{9}+i\cos\dfrac{2\pi}{9}}{1+\sin\dfrac{2\pi}{9}-i\cos\dfrac{2\pi}{9}}\right)^3 is
  1. A.
    12(1−i3)\dfrac{1}{2}\left(1-i\sqrt{3}\right)
  2. B.
    −12(3−i)-\dfrac{1}{2}\left(\sqrt{3}-i\right)
  3. C.
    −12(1−i3)-\dfrac{1}{2}\left(1-i\sqrt{3}\right)
  4. D.
    12(3+i)\dfrac{1}{2}\left(\sqrt{3}+i\right)
Show answer and solution

Correct option: B

Solution

The key is to rewrite the expression inside the brackets. Note that sin(2π/9) = cos(π/2 - 2π/9) = cos(5π/18) and cos(2π/9) = sin(5π/18), so the numerator becomes 1 + cos(5π/18) + i·sin(5π/18) and denominator is 1 + cos(5π/18) - i·sin(5π/18).

Using the identity 1 + cosθ + i·sinθ = 2cos(θ/2)·[cos(θ/2) + i·sin(θ/2)] and similarly for denominator, the fraction equals [cos(θ/2) + i·sin(θ/2)] / [cos(θ/2) - i·sin(θ/2)] = e^(iθ) = cos θ + i·sin θ, where θ = 5π/18.

So the base equals cos(5π/18) + i·sin(5π/18), and cubing gives cos(5π/6) + i·sin(5π/6).

cos(5π/6) = -√3/2 and sin(5π/6) = 1/2, giving -√3/2 + i/2 = -(1/2)(√3 - i), which is option 2.

[JEE Main 2023 · 24 Jan · Shift 2]

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