Question

If $A,B,C$ be the angles of a triangle, then prove that
$\left|\begin{array}{ccc}{e}^{2iA}& e^{-iC}& e^{-iB}\\ e^{-iC}& e^{2iB}& e^{-iA}\\ e^{-iB}& e^{-iA}& e^{2iC}\end{array}\right|$ is purely real.

Answer 1

This question is based on the following :
$A+B+C=\pi$, $e^{\pi i}=\cos \pi +i\sin \pi =-1$, $e^{-\pi i}=-1$
$e^{-i(B+C)}=e^{i(\pi -A)}=e^{\pi i}$. $e^{-iA}=-e^{-iA}$
$\therefore$ $e^{-i(B+C)}=-e^{iA}$
Take $e^{iA}$, $e^{iB}$ and $e^{iC}$ common from $R_1$, $R_2$ and $R_3$ respectively.
$\therefore$ $\Delta =e^{i(A+B+C)}\left|\begin{array}{ccc}{e}^{iA}& e^{-i(A+C)}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{iB}& e^{-i(B+A)}\\ e^{-i(B+C)}& e^{-i(C+A)}& e^{iC}\end{array}\right|$
$=-1\left|\begin{array}{ccc}{e}^{iA}& {-e}^{iB}& {-e}^{iC}\\ {-e}^{iA}& e^{iB}& {-e}^{iC}\\ {-e}^{iA}& {-e}^{iB}& e^{iC}\end{array}\right|$,by $\left(2\right)$
Take $e^{iA}$, $e^{iB}$ and $e^{iC}$ common from $C_1$, $C_2$ and $C_3$ and again put $e^{i(A+B+C)}=e^{i\pi }=-1$.
$\therefore$ $\Delta =(-1)(-1)\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right|$
Now make two zeros and expand $\Delta =-4$ which is purely
real.