Question

Answer the following:

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

Answer 1

Let `sqrt(2 + 2sqrt(3)"i"` = a + bi, where a, b ∈ R.

Squaring on both sides, we get

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