Question

Which of the following are square roots of —8 + 8i/3? Check all that apply.

Answer 1

Answer:

Options (2) and (3)

Step-by-step explanation:

Let, $\sqrt{-8+8i\sqrt{3}}=(a+bi)$

$(\sqrt{-8+8i\sqrt{3}})^2=(a+bi)^2$

-8 + 8i√3 = a² + b²i² + 2abi

-8 + 8i√3 = a² - b² + 2abi

By comparing both the sides of the equation,

a² - b² = -8 -------(1)

2ab = 8√3

ab = 4√3 ----------(2)

a = $\frac{4\sqrt{3}}{b}$

By substituting the value of a in equation (1),

$(\frac{4\sqrt{3}}{b})^2-b^2=-8$

$\frac{48}{b^2}-b^2=-8$

48 - b⁴ = -8b²

b⁴ - 8b² - 48 = 0

b⁴ - 12b² + 4b² - 48 = 0

b²(b² - 12) + 4(b² - 12) = 0

(b² + 4)(b² - 12) = 0

b² + 4 = 0 ⇒ b = ±√-4

                     b = ± 2i

b² - 12 = 0 ⇒ b = ±2√3

Since, a = $\frac{4\sqrt{3}}{b}$

For b = ±2i,

a = $\frac{4\sqrt{3}}{\pm2i}$

  = $\pm\frac{2i\sqrt{3}}{(-1)}$

  = $\mp 2i\sqrt{3}$

But a is real therefore, a ≠ ±2i√3.

For b = ±2√3

a = $\frac{4\sqrt{3}}{\pm 2\sqrt{3}}$

a = ±2

Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)

Options (2) and (3) are the correct options.