Question

The complex number $\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugate to each other for :`

• A. $x=n\pi$
• B. $x=(n+\frac{1}{2})\pi$
• C. $x=0$
• D. No value of $x$.

Answer 1

The correct option is D No value of $x$.

Let $z_1=\sin x+i\cos 2x$ and $z_2=\cos x-i\sin 2x$

Given, numbers are conjugate to each other,

$\therefore \overline{z_1}=z_2$ or $\overline{z_2}=z_1$

$\overline{z_1}=z_2$
$\Rightarrow \sin x-i\cos 2x=\cos x-i\sin 2x$
Equating real and imaginary parts, we get
$\sin x=\cos x$ and $\cos 2x=\sin 2x$
$\therefore \tan x=1$
$\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},...$ $\left(i\right)$
and $\tan 2x=1$
$\Rightarrow 2x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4},...$
$\Rightarrow x=\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},...$ $\left(ii\right)$
None of the values of $x$ are common in Eqs. $\left(i\right)$ and $\left(ii\right)$

Hence, $z_1$ and $z_2$ are conjugate to each other for no values of $x$.