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Question
Answer the following:
show that `((1 + "i")/sqrt(2))^8 + ((1 - "i")/sqrt(2))^8` = 2
Solution
`((1 + "i")/sqrt(2))^2 = (1 + 2"i" + "i"^2)/2 = (1 + 2"i" - 1)/2` = i
∴ `((1 + "i")/sqrt(2))^8 = [((1 + "i")/sqrt(2))^2]^4` = i4 = 1 .......(i)
Also, `((1 - "i")/sqrt(2))^2 = (1 - 2"i" + "i"^2)/2 = (1 - 2"i" - 1)/2` = – i
∴ `((1 - "i")/sqrt(2))^8 = [((1 - "i")/sqrt(2))^2]^4`
= (– i)4 = (– 1)4 × (i)4
= 1 × i4
= 1 ........(ii)
Adding (i) and (ii), we get
`((1 + "i")/sqrt(2))^8 + ((1 - "i")/sqrt(2))^8` = 1 + 1 = 2
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