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प्रश्न
If ω is the cube root of unity then find the value of `((-1 + "i"sqrt(3))/2)^18 + ((-1 - "i"sqrt(3))/2)^18`
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उत्तर १
`((-1 + "i"sqrt(3))/2)^3`
`=((-1)^3 + 3(-1)^2("i"sqrt(3)) + 3(-1)("i"sqrt(3))^2 + ("i"sqrt(3))^3)/8`
= `(-1 + 3(1)("i"sqrt(3)) - 3(3"i"^2) + 3sqrt(3)"i"^3)/8`
= `(-1 + 3sqrt(3)"i" + 9 - 3sqrt(3)"i")/8` ...[∵ i2 = – 1, i3 = – i]
= `8/8`
= 1 .......(1)
Also, `((-1 - "i"sqrt(3))/2)^3`
`=((-1)^3 - 3(-1)^2("i"sqrt(3)) + 3(-1)("i"sqrt(3))^2 - ("i"sqrt(3))^3)/8`
= `(-1 - 3(1)("i"sqrt(3)) - 3(3"i"^2) - 3sqrt(3)"i"^3)/8`
= `(-1 - 3sqrt(3)"i" + 9 + 3sqrt(3)"i")/8` ...[∵ i2 = – 1, i3 = – i]
= `8/8`
= 1 .........(2)
∴ `((-1 + "i"sqrt(3))/2)^18 + [((-1 - "i"sqrt(3))/2)^2]^18`
= `[((-1 + "i"sqrt(3))/2)^3]^6 + [((-1 - "i"sqrt(3))/2)^36]`
= `[((-1 + "i"sqrt(3))/2)^3]^6 + [((-1 - "i"sqrt(3))/2)^3]^12`
= (1)6 + (1)12 ........[By (1) and (2)]
= 1 + 1
= 2
उत्तर २
If ω is the complex cube root of unity, then
ω3 = 1, ω = `(-1 + isqrt3)/2` and ω2 = `(-1 - isqrt3)/2`
Consider,
`((-1 + isqrt3)/2)^18` + `((-1 - isqrt3)/2)^18`
Given Expression = ω18 + (ω2)18
= ω18 + ω36
= (ω3)6 + (ω3)12
= (1)6 + (1)12
= 1 + 1
= 2
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