Question

How do you cube `1+sqrt3i`?

Answer 1

Use Binomial expansion or De Moivre's formula...

Explanation:

Method 1

Binomial expansion...

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

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

`=1+3sqrt(3)i+9i^2+3sqrt(3)i^3`

`=1+3sqrt(3)i-9-3sqrt(3)i`

`=(1-9)+(3sqrt(3)-3sqrt(3))i`

`=-8`

Method 2

Using De Moivre's formula...

`1+sqrt(3)i = 2(1/2 + sqrt(3)/2i) = 2(cos(pi/3)+sin(pi/3)i)`

So

`(1+sqrt(3)i)^3 = (2(cos(pi/3)+i sin(pi/3)))^3 = 2^3(cos(pi/3)+i sin(pi/3))^3`

`=8 (cos pi + i sin pi) = 8 * -1 = -8`