Question

Find the square root of the following complex number:

`1 + 4sqrt(3)"i"`

Answer 1

Let `sqrt(1 + 4sqrt(3)"i")` = a + bi, where a, b ∈ R

Squaring on both sides, we get

`1 + 4sqrt(3)"i"` = (a + bi)² = a² + b²i² + 2abi

∴ `1 + 4sqrt(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 –  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")`