If A-B-C-ei-cos-isin-and z- - e2iA e-iC e-iB e-iC e2iB e-iA e-iB e-iA e2iC - then

The correct options are B Re(z=) 4 C Im(z)=0 | e^2iA amp;e^ iC #160; amp;e^ iB e^ iC #160; amp;e^2iB #160; amp;e^ iA e^ iB ...

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Adjoint of a Matrix

If A+B+C=π,...

Question

If $A + B + C = π, e^{i θ} = cos θ + i sin θ$ and z= $∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{2 i A} & e^{- i C} & e^{- i B} e^{- i C} & e^{2 i B} & e^{- i A} e^{- i B} & e^{- i A} & e^{2 i C} \end{matrix} ∣ ∣ ∣ ∣ ∣$ then

R e (z) = 4

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I m (z) = 0

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R e (z =) - 4

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I m (z) = - 1

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Solution

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Similar questions

A, B, C

∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{2 i A} & e^{- i C} & e^{- i B} e^{- i C} & e^{2 i B} & e^{- i A} e^{- i B} & e^{- i A} & e^{2 i C} \end{matrix} ∣ ∣ ∣ ∣ ∣

z = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{2 i A} & e^{- i C} & e^{- i B} e^{- i C} & e^{2 i B} & e^{- i A} e^{- i B} & e^{- i A} & e^{2 i C} \end{matrix} ∣ ∣ ∣ ∣ ∣

A + B + C = π

e^{i θ} = cos θ + i sin θ

A, B, C

∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{2 i A} & e^{- i C} & e^{- i B} e^{- i C} & e^{2 i B} & e^{- i A} e^{- i B} & e^{- i A} & e^{2 i C} \end{matrix} ∣ ∣ ∣ ∣ ∣

sin θ = (\frac{z - 1}{i}),

z

θ

z = {(\frac{\sqrt{3}}{2} + \frac{i}{2})}^{5} + {(\frac{\sqrt{3}}{2} - \frac{i}{2})}^{5},

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