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Question
Show that `(1/sqrt(2) + "i"/sqrt(2))^10 + (1/sqrt(2) - "i"/sqrt(2))^10` = 0
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Solution
`(1/sqrt(2) + "i"/sqrt(2))^2 = 1/2 + 2(1/sqrt(2)) ("i"/sqrt(2)) + "i"^2/2`
= `1/2 + "i" - 1/2` = i
∴ `(1/sqrt(2) + "i"/sqrt(2))^10 = [(1/sqrt(2) + "i"/sqrt(2))^2]^5 = i^5 = i^4 * i = 1*i = i ...(i)`
Also, `(1/sqrt(2) - "i"/sqrt(2))^2 = 1/2 - 2(1/sqrt(2)) ("i"/sqrt(2)) + "i"^2/2`
= `1/2 - "i" - 1/2` = – i
∴ `(1/sqrt(2) - "i"/sqrt(2))^10 = [(1/sqrt(2) - "i"/sqrt(2))^2]^5 =(-i)^5 = i^4 * (-i) = 1*(-i) = -i ...(ii)`
Adding (i) and (ii), we get
`(1/sqrt(2) + "i"/sqrt(2))^10 + (1/sqrt(2) - "i"/sqrt(2))^10` = i – i = 0
Hence proved.
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