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Question: The complex numbers \(\sin x + i\cos{}2x\) and \(\cos x - i\sin{}2x\) are conjugate to each other fo...

The complex numbers sin⁡x+icos⁡2x\sin x + i\cos{}2x and cos⁡x−isin⁡2x\cos x - i\sin{}2x are conjugate to each other for

A

x=nπx = n\pi

B

x=(n+12)πx = \left( n + \frac{1}{2} \right)\pi

C

x=0x = 0

D

No value of x

Answer

No value of x

Explanation

Solution

Sol.sin⁡x+icos⁡2x\sin x + i\cos 2xandcos⁡x−isin⁡2x\cos x - i\sin{}2xare conjugate to each other if sin⁡x=cos⁡xandcos⁡2x=sin⁡2x\sin x = \cos x\text{and}\cos 2x = \sin 2x

or tan⁡x=1⇒x=π4,5π4,9π4,........\tan x = 1 \Rightarrow x = \frac{\pi}{4},\frac{5\pi}{4},\frac{9\pi}{4},........ (i) and

tan⁡2x=1⇒2x=π4,5π4,9π4,........\tan 2x = 1 \Rightarrow 2x = \frac{\pi}{4},\frac{5\pi}{4},\frac{9\pi}{4},........ or x=π8,5π8,9π8,.......x = \frac{\pi}{8},\frac{5\pi}{8},\frac{9\pi}{8},....... (ii) There exists no value of x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate.