How do you cube #1+sqrt3i#?

1 Answer
Oct 19, 2015

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#