Answer:
Options (2) and (3)
Step-by-step explanation:
Let, [tex]\sqrt{-8+8i\sqrt{3}}=(a+bi)[/tex]
[tex](\sqrt{-8+8i\sqrt{3}})^2=(a+bi)^2[/tex]
-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 = [tex]\frac{4\sqrt{3}}{b}[/tex]
By substituting the value of a in equation (1),
[tex](\frac{4\sqrt{3}}{b})^2-b^2=-8[/tex]
[tex]\frac{48}{b^2}-b^2=-8[/tex]
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 = [tex]\frac{4\sqrt{3}}{b}[/tex]
For b = ±2i,
a = [tex]\frac{4\sqrt{3}}{\pm2i}[/tex]
= [tex]\pm\frac{2i\sqrt{3}}{(-1)}[/tex]
= [tex]\mp 2i\sqrt{3}[/tex]
But a is real therefore, a ≠ ±2i√3.
For b = ±2√3
a = [tex]\frac{4\sqrt{3}}{\pm 2\sqrt{3}}[/tex]
a = ±2
Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)
Options (2) and (3) are the correct options.
