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Show that (-1+3i)3 is a real number

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Question

Show that `(-1 + sqrt(3)"i")^3` is a real number

Sum
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Solution

`(-1 + sqrt(3)"i")^3`

`= (-1)^3 + 3(-1)^2 (sqrt(3)"i") + 3(-1) (sqrt(3)"i")^2 + (sqrt(3"i"))^3`   ...[(a + b)3 = a3 + 3a2b + 3ab2 + b3]

= `-1 + 3sqrt(3)"i" - 3(3"i"^2) + 3sqrt(3)"i"^3`

= `-1 + 3sqrt(3)"i" - 3(-3) - 3sqrt(3)"i"`   ...[∵ i2 = – 1, i3 = – i]

= `-1 + 3sqrt(3)"i" + 9 - 3sqrt(3)"i"`   ...[∵ i2 = – 1, i3 = – i]

= – 1 + 9

= 8, which is a real number

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Chapter 1: Complex Numbers - Exercise 1.1 [Page 6]

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