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#