Question

Find the square root of the following complex number: 

`2(1 - sqrt(3)"i")`

Answer 1

Let `sqrt(2(1 -sqrt3"i")` = a + bi, where a, b ∈ R

Squaring on both sides, we get

`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 - (-(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(1 - sqrt(3)"i")) = ± (sqrt(3) - "i")`