Question

Find the square root of the following complex numbers: `2(1 - sqrt(3)  "i")`

Answer 1

Let `sqrt(2(1 - sqrt(3) "i")`= a + bi, where a, b ∈ R
Squaring on both sides, we get
`2(1 - sqrt(3) "i")` = (a + bi)²
∴ `2(1 - sqrt(3) "i")` = a² + b²i² + 2abi
∴ `2 - 2sqrt(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 - (- sqrt3/"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(1 - sqrt(3)"i")) = ± (sqrt(3) - "i")`.