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z – z¯ = 6i ( z – z¯ = 6i)
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(1 + 2i) 2 (1 – i)
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(1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)
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(1 – ω – ω2)3 + (1 – ω + ω2)3
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(1 + ω – ω2) 6 = 64
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If ω is the complex cube root of unity, show that (i) (2 – ω)(2 – ω2) = 7
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18i complex number
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i131 + i49 complex number
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2x + i9 y(2 + i) = x i7 + 10 i16
asked Sep 11, 2022 in Complex Numbers by Doubtly (98.9k points)
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