Question

Simplify the expression (3 – 3i)3 by performing operations with pure imaginary numbers and complex numbers. (Provide an explanation as I am terrible at calculus and need major assistance, thanks!)

A) –27 + 216i

B) –27 + 54i

C) –54 – 216i

D) –54 – 54i

Answer 1

D

Explanation:

Given:
`z=3-3i`

Find:
`z^3=(3-3i)^3`

Solution:

Use the formula

`(a-b)^3=a^3-3a^2b+3ab^2-b^3`

Hence,

`(3-3i)^3\\ \\=3^3-3\cdot 3^2\cdot(3i)+3\cdot 3\cdot (3i)^2-(3i)^3\\ \\=27-81i+9\cdot 9i^2-3^3i^3\\ \\=27-81i+81i^2-27i^3`

Now remind that

`i^2=-1,`

then

`i^3=i^2\cdot i=-1\cdot i=-i`

Substitute:

`(3-3i)^3\\ \\=27-81i+81\cdot (-1)-27\cdot (-i)\\ \\=27-81i-81+27i\\ \\=27-81-81i+27i\\ \\=-54-54i`