मराठी

If ω is a complex cube root of unity, then prove the following: (a + b) + (aω + bω2) + (aω2 + bω) = 0.

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प्रश्न

If ω is a complex cube root of unity, then prove the following:  (a + b) + (aω + bω2) + (aω2 + bω) = 0.

बेरीज
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उत्तर

ω is a 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. = (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 = R.H.S.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Complex Numbers - EXERCISE 3.3 [पृष्ठ ४२]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 11 Maharashtra State Board
पाठ 3 Complex Numbers
EXERCISE 3.3 | Q 5) ii) | पृष्ठ ४२

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