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Answer the following: show that ii(1+i2)8+(1-i2)8 = 2

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Question

Answer the following:

show that `((1 + "i")/sqrt(2))^8 + ((1 - "i")/sqrt(2))^8` = 2

Sum
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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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Chapter 1: Complex Numbers - Miscellaneous Exercise 1.2 [Page 22]

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