Question

Find the square root of: `2 + 2 sqrt(3)"i"`

Answer 1

Let `sqrt(2 + 2sqrt(3)"i")` = a + bi, where a, b ∈ R.
Squaring on both sides, we get
`2 + 2 sqrt(3)"i"` = a² + b²i² + 2abi
∴ `2 + 2 sqrt(3)"i"` = a² – b² + 2abi      ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 2 and 2ab = `2sqrt(3)`

∴ a² – b² = 2 and b = `sqrt(3)/"a"`

∴ `"a"^2 - (sqrt(3)/"a")^2` = 2

∴ `"a"^2 - 3/"a"^2` = 2

∴ a⁴ – 3 = 2a²
∴ a⁴ – 2a² – 3 = 0
∴ (a² – 3)(a² + 1) = 0
∴ a² = 3 or a² = – 1
But a ∈ R
∴ a² ≠ – 1
∴ a²  = 3
∴ a = ± `sqrt(3)`

When a = `sqrt(3), "b" = sqrt(3)/sqrt(3)` = 1

When a = `-sqrt(3), "b" = sqrt(3)/-sqrt(3)` = – 1

∴ `sqrt(2 + 2sqrt(3)"i") = ± (sqrt(3) + "i")`.