Question

If $e^{ix}=\cos x+i\sin x$ and
$x+iy=\left|\begin{array}{ccc}1& e^{\pi i/4}& e^{\pi i/3}\\ e^{-\pi i/4}& 1& e^{2\pi i/3}\\ e^{-\pi i/3}& e^{-2\pi i/3}& e^{-2\pi i}\end{array}\right|$, then

• A. $x=-1,y=\sqrt{2}$
• B. $x=1,y=-\sqrt{2}$
• C. $x=-\sqrt{2},y=\sqrt{2}$
• D. none of these

Answer 1

The correct option is D none of these

$x+iy=\left|\begin{array}{ccc}1& e^{\pi i/4}& e^{\pi i/3}\\ e^{-\pi i/4}& 1& e^{2\pi i/3}\\ e^{-\pi i/3}& e^{-2\pi i/3}& e^{-2\pi i}\end{array}\right|$

Taking conjugate

$x-iy=\left|\begin{array}{ccc}1& e^{-\pi i/4}& e^{-\pi i/3}\\ e^{\pi i/4}& 1& e^{-2\pi i/3}\\ e^{\pi i/3}& e^{2\pi i/3}& e^{2\pi i}\end{array}\right|$

Changing rows by column

$x-iy=\left|\begin{array}{ccc}1& e^{\pi i/4}& e^{\pi i/3}\\ e^{-\pi i/4}& 1& e^{2\pi i/3}\\ e^{-\pi i/3}& e^{-2\pi i/3}& e^{-2\pi i}\end{array}\right|$

Therefore $x+iy=x-iy\Rightarrow y=0$