Question

If $a=\cos \theta +i\sin \theta ,b=\cos 2\theta -i\sin 2\theta ,c=\cos 3\theta +i\sin 3\theta$, where $i=\sqrt{-1}$ and if
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$, then

• A. $\theta =2k\pi ,kϵZ$
• B. $\theta =(2k+1)\pi ,kϵZ$
• C. $\theta =(4k+1)\pi ,kϵZ$
• D. none of these

Answer 1

The correct option is A $\theta =2k\pi ,kϵZ$

$\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
$=-(a^3+b^3+c^3-3abc)$
$=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=-\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
$\Rightarrow a+b+c=0$ or $a=b=c$
If $a+b+c=0$, we have
$\cos \theta +\cos 2\theta +\cos 3\theta =0$ and $\sin \theta -\sin 2\theta +\sin 3\theta =0$
$\Rightarrow \cos 2\theta (2\cos \theta +1)=0$ and $\sin 2\theta (2\cos \theta -1)=0\cdots \left(1\right)$
which is not possible as $\cos 2\theta =0$ gives $\sin 2\theta \ne 0,\cos \theta \ne 1/2$
Therefore, Eq.$\left(1\right)$ does not hold simultaneously.
$\therefore a+b+c\ne 0$
$\therefore a=b=c$
or
$e^{i\theta }=e^{-2i\theta }=e^{3i\theta }$
which is satisfied only by $e^{i\theta }=1$, i.e.,$\cos \theta =1,\sin \theta =0$ so $\theta =2k\pi ,kϵZ.$