If A + B + C = π, eiθ = cosθ + isinθ and then

Engineering

Mathematics

Introduction to Determinants

Question

If A + B + C = π, e^iθ = cosθ + isinθ and $z = ∣ \begin{array}{ccc} e^{2 iA} & e^{- iC} & e^{- iB} \\ e^{- iC} & e^{2 iB} & e^{- iA} \\ e^{- iB} & e^{- iA} & e^{2 iC} \end{array} ∣$ then

Re(z) = 4

lm(z) = 0

Re(z) = – 4

lm(z) = – 1

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∣ \begin{array}{ccc} e^{2 iA} & e^{- iC} & e^{- iB} \\ e^{- iC} & e^{2 iB} & e^{- iA} \\ e^{- iB} & e^{- iA} & e^{2 iC} \end{array} ∣

If you expand it,
= e^2iA(e^2i(B+C) – e^–2iA) – e^–iC(e^iC – e^–i(A+B)) + e^–iB(e^–i(C+A) – e^iB)

= e^{2i(A + B + C)} – 1 – 1 + e^{–i(A + B + C)} + e^{–i(A + B + C)} – 1
= e^2iπ + 2e^–iπ – 3

= 1 – 2 – 3
= – 4

So, it has lm(z) = 0 and Re(z) = – 4

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