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If ω is a complex cube root of unity, show that (a + b) + (aω + bω2) + (aω2 + bω) = 0 - Mathematics and Statistics

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प्रश्न

If ω is a complex cube root of unity, show that (a + b) + (aω + bω2) + (aω2 + bω) = 0

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उत्तर

ω is a complex cube root of unity.

∴ ω3 = 1 and 1 + ω + ω2 = 0.

∴ ω + ω2 = – 1, 1 + ω = – ω2 and 1 + ω2 = – ω.

(a + b) + (aω + bω2) + (aω2 + bω)

= a + b + aω + bω2 + aω2 + bω

= (a + aω + aω2) + (b + bω + bω2)

= a(1 + ω + ω2) + b(1 + ω + ω2)

= a(0) + b(0)

= 0

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पाठ 1: Complex Numbers - Exercise 1.4 [पृष्ठ २०]

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