If a = e i α , b = e i β and c = e i γ and b a + c b + a c = 1 , find c y c ∑ cos ( α − β ) .
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How ∑ c y c sin ( α − β ) = 0 ?
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2 Complex Numbers are equal if their real and imaginary parts are equal.
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What are you tryin to say? Please elaborate.
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@Sahil Silare – Let { z 1 = x 1 + i y 1 z 2 = x 2 + i y 2 , z 1 = z 2 if and only if x 1 = x 2 or ℜ ( z 1 ) = ℜ ( z 2 ) and y 1 = y 2 or ℑ ( z 1 ) = ℑ ( z 2 ) .
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b a + c b + a c e i β e i α + e i γ e i β + e i α e i γ e i ( α − β ) + e i ( β − γ ) + e i ( γ − α ) cos ( α − β ) + i sin ( α − β ) + cos ( β − γ ) + i sin ( β − γ ) + cos ( γ − α ) + i sin ( γ − α ) c y c ∑ cos ( α − β ) + i c y c ∑ sin ( α − β ) ⟹ c y c ∑ cos ( α − β ) = 1 = 1 = 1 = 1 = 1 + i 0 = 1 Using Euler’s formula: e i θ = cos θ + i sin θ Equating the real part