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Question
Answer the following:
Show that `(1/sqrt(2) + "i"/sqrt(2))^10 + (1/sqrt(2) - "i"/sqrt(2))^10` = 0
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`
= i5 = i4.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
= i4(– 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
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