SOLUTION: I'm having trouble with a question in my lesson. Could I have some help? (-1 + [SQRT(3)]i)^3 These are the options: A) 2 + 2[SQRT(3)]i B) -2 - 2[SQRT(3)]i C) 0 D) -

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I'm having trouble with a question in my lesson. Could I have some help? (-1 + [SQRT(3)]i)^3 These are the options: A) 2 + 2[SQRT(3)]i B) -2 - 2[SQRT(3)]i C) 0 D) -      Log On


   

Question 740092: I'm having trouble with a question in my lesson. Could I have some help?
(-1 + [SQRT(3)]i)^3
These are the options:
A) 2 + 2[SQRT(3)]i
B) -2 - 2[SQRT(3)]i
C) 0
D) -8
E) -8 + 8[SQRT(3)]i
F) 8
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You can calculate %28-1%2Bsqrt%283%29i%29%5E3 as you would calculate the cube of any binomial, using the formula
%28a%2Bb%29%5E3=a%5E3%2B3a%5E2b%2B3ab%5E2%2Bb%5E3


If you already learned about the polar form of complex numbers, you could transform the number into its polar form, if that made it easier for you.
-1%2Bsqrt%283%29i=2%28cos%28120%5Eo%29%2Bi%2Asin%28120%5Eo%29%29 has
absolute value (or modulus): r=2+=sqrt%28%28-1%29%5E2%2B%28sqrt%283%29%29%5E2%29
argument: theta=120%5Eo
Multiplication and powers are easier in polar form.
To multiply, you just multiply the absolute values and add the arguments.
So