JEE PYQ: Complex Numbers Question 14
Question 14 - 2020 (02 Sep Shift 1)
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:
(1) $\frac{1}{2}(1-i\sqrt{3})$
(2) $\frac{1}{2}(\sqrt{3}-i)$
(3) $-\frac{1}{2}(\sqrt{3}-i)$
(4) $-\frac{1}{2}(1-i\sqrt{3})$
Show Answer
Answer: (3)
Solution
$\sin\frac{2\pi}{9} = \cos\frac{\pi}{18-}$… The expression simplifies using $\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} = e^{i(\pi/2-\theta)}$ type identity. Result: $-\frac{1}{2}(\sqrt{3}-i)$.
BETA
