Question

Find the square root of the following complex numbers: 1 + 4 `sqrt(3)  "i"`

Answer 1

Let `sqrt(1 + 4sqrt(3) "i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
1 + 4 `sqrt(3)  "i"` = (a + bi)²
∴ 1 + 4 `sqrt(3)  "i"` = a² + b²i² + 2abi
∴ 1 + 4 `sqrt(3)  "i"` = (a² – b²) + 2abi       ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 1 and 2ab = `4sqrt(3)`

∴ a² – b² = 1 and b = `(2sqrt(3))/"a"`

∴ `"a"^2 - ((2sqrt(3))/"a")^2` = 1

∴ `"a"^2 - 12/"a"^2` = 1

∴ a⁴ – 12 = a²
∴ a⁴ – a² – 12 = 0
∴ (a² – 4)(a² + 3) = 0
∴ a² = 4 or a² = – 3
But a ∈ R
∴ a² ≠ – 3
∴ a² = 4
∴ a = ± 2
When a = 2, b = `(2sqrt(3))/2 = sqrt(3)`

When a = – 2, b = `(2sqrt(3))/-2 = -sqrt(3)`

∴ `sqrt(1 + 4sqrt(3)"i") = ± (2 + sqrt(3)"i")`