Complex Numbers 2 Question 17
17. The complex numbers $\sin x+i \cos 2 x$ and $\cos x-i \sin 2 x$ are conjugate to each other, for
(a) $x=n \pi$
(b) $x=0$
(c) $x=(n+1 / 2) \pi$
(d) no value of $x$
(1988, 2M)
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Answer:
Correct Answer: 17. (a, c, d)
Solution:
- Since, $(\overline{\sin x+i \cos 2 x})=\cos x-i \sin 2 x$
$\Rightarrow \quad \sin x-i \cos 2 x=\cos x-i \sin 2 x$
$\Rightarrow \quad \sin x=\cos x$ and $\cos 2 x=\sin 2 x$
$\Rightarrow \quad \tan x=1$ and $\tan 2 x=1$
$\Rightarrow \quad x=\pi / 4$ and $x=\pi / 8$ which is not possible at same time.
Hence, no solution exists.
