The value of $\left(\frac{1+\sin \frac{2\pi}{9}+i\cos \frac{2\pi}{9}}{1+\sin \frac{2\pi}{9}-i\cos \frac{2\pi}{9}}\right)^{3}$ is

(C) $\frac{-1}{2}(\sqrt{3}-i)$

(A) $\frac{-1}{2}(1-i\sqrt{3})$

(B) $\frac{1}{2}(1-i\sqrt{3})$

(D) $\frac{1}{2}(\sqrt{3}+i)$

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Complex Numbers

The value of

(\frac{1 + s i n \frac{2 π}{9} + i c o s \frac{2 π}{9}}{1 + s i n \frac{2 π}{9} - i c o s \frac{2 π}{9}})^{3}

Answer: (C)

Let

z = sin \frac{2 π}{9} + i cos \frac{2 π}{9}

, which has unit modulus, so

\overset{z}{ˉ} = \frac{1}{z}

. The denominator is

1 + \overset{z}{ˉ} = 1 + \frac{1}{z} = \frac{z + 1}{z}

, hence

\frac{1 + z}{1 + z ˉ} = z

. Writing

z = cos (\frac{π}{2} - \frac{2 π}{9}) + i sin (\frac{π}{2} - \frac{2 π}{9}) = e^{i (π /2 - 2 π /9)}

, the cube is

z^{3} = e^{i (3 π /2 - 2 π /3)} = e^{i 5 π /6} = cos \frac{5 π}{6} + i sin \frac{5 π}{6} = - \frac{3}{2} + \frac{i}{2}

. The value is

- \frac{1}{2} (3 - i)

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Single correct option