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If ω is a complex cube root of unity, show that (1 + ω − ω2)6 = 64

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प्रश्न

If ω is a complex cube root of unity, show that (1 + ω − ω2)6 = 64

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उत्तर

ω is the complex cube root of unity

∴ ω3 = 1 and 1 + ω + ω2 = 0

Also, 1 + ω2 = − ω, 1 + ω = − ω2 and ω + ω2 = − 1

L.H.S. = (1 + ω − ω2)6

= [(1 + ω) − ω2]6

= (−ω2 − ω2)6

= (−2ω2)6

= 64.ω12

= 64(ω3)4

= 64(1)4

= 64

= R.H.S.

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

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