Prove that determinant involving angles of triangle is purely real

Question

If $A, B, C$ are the angles of a triangle, prove that

$$ \begin{vmatrix} e^{2iA} & e^{-iC} & e^{-iB} \\\ e^{-iC} & e^{2iB} & e^{-iA} \\\ e^{-iB} & e^{-iA} & e^{2iC} \end{vmatrix}$$

is purely real.

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I tried solving this question, even by expanding the matrix directly. But in the end, I get terms of the form $ e^{i(f(A, B, C))} $ from which I am unable to converge. Is this because I haven't studied complex numbers deeply yet?

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Note: This question originates from SK Goyal's book in Algebra for JEE and other competitive exams in India.

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Answer 1

Directly expanding the determinant, we get $$ e^{2iA + 2iB + 2iC} + 2e^{-iA -iB -iC} - 3$$ and since $A + B + C = \pi$, this determinant is equal to $$ e^{2 \pi i} + 2 e^{-i \pi} - 3 = 1 - 2 - 3 = -4.$$

Answer 2

$$ \det\begin{bmatrix} e^{2iA} & e^{-iC} & e^{-iB} \\\ e^{-iC} & e^{2iB} & e^{-iA} \\\ e^{-iB} & e^{-iA} & e^{2iC} \end{bmatrix} = e^{-i(A+B+C)}\left(2-3e^{i(A+B+C)}+e^{3i(A+B+C)}\right)$$

but $A+B+C = \pi$ then $ e^{-i(A+B+C)}\left(2-3e^{i(A+B+C)}+e^{3i(A+B+C)}\right) = -4$