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If ω is a complex cube root of unity, show that ((a + bomega + comega^2))/("c" + aomega + bomega^2) = omega^2

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प्रश्न

If ω is a complex cube root of unity, show that `((a + bomega + comega^2))/("c" + aomega + bomega^2) = omega^2`.

योग
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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 + bomega + comega^2)/(c + aomega + bomega^2)`

= `(aomega^3 + bomega^4  + comega^2)/(c + aomega + bomega^2)     ...[∵ omega^3 = 1, omega^4 = omega]`

= `(omega^2(c + aomega + bomega^2))/(c + aomega + bomega^2)`

= ω2

= R.H.S.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Complex Numbers - EXERCISE 3.3 [पृष्ठ ४२]

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

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