Question

Show that √3 + i√3 and -√3 - i√3 are roots of 6i. Find the square of each number.

a. (√3 + i√3)^2 = 6i, (-√3 - i√3)^2 = -6i
b. (√3 + i√3)^2 = -6i, (-√3 - i√3)^2 = 6i
c. (√3 + i√3)^2 = -6i, (-√3 - i√3)^2 = -6i
d. (√3 + i√3)^2 = 6i, (-√3 - i√3)^2 = 6i

Answer 1

By squaring both √3 + i√3 and -√3 - i√3, it is confirmed that their squares are 6i and -6i, respectively, showing that they are roots of 6i.

Step-by-step explanation:

To show that √3 + i√3 and -√3 - i√3 are roots of 6i, we need to find the square of each number.

Let's calculate the square of the first number: (√3 + i√3)^2.

• ((√3)^2 + 2(√3)(i√3) + (i√3)^2)
• (3 + 2i3 + 3i^2)
• (3 + 6i - 3)
• (6i)

The square of (√3 + i√3) is indeed 6i, which confirms it as a root.

For the second number: (-√3 - i√3)^2.

• ((-√3)^2 - 2(√3)(i√3) + (-i√3)^2)
• (3 - 2i3 + 3i^2)
• (3 - 6i - 3)
• (-6i)

The square of (-√3 - i√3) is -6i, which confirms it as the other root.

Therefore, the correct answer to the question is a. (√3 + i√3)^2 = 6i, (-√3 - i√3)^2 = -6i.