Question

The complex numbers $sinx+icos2x$ and $cosx-isin2x$ are conjugate to each other for

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

Answer 1

The correct option is D no value of x

$z=e^{i\theta }$
$=cosx-isin2x$
$z_2=\overline{z}=e^{-i\theta }$
$=cosx+isin2x$
$=sinx+icos2x$ ....(given)
Comparing coefficients, we get
$cosx=sinx$ and $cos2x=sin2x$
$x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}...$ and $x=\frac{\pi }{8},\frac{5\pi }{8}..$
The intersection of the solution sets is $\varphi$.
Therefore, there are no values of $x$ satisfying the given conditions.
Hence, option 'D' is correct.