Answer 1

Let `sqrt(3 + 2sqrt(10)"i")` = a + bi, where a, b ∈ R

Squaring on both sides, we get

`3 + 2sqrt(10)"i"` = a² + b²i² + 2abi

∴ `3 + 2sqrt(10)"i"` = (a² – b²) + 2abi   ...[∵ i² = – 1]

Equating real and imaginary parts, we get

a² – b² = 3 and 2ab = `2sqrt(10)`

∴ a² – b² = 3 and b = `(sqrt(10))/"a"`

∴ `"a"^2 - ((sqrt(10))/"a")^2` = 3

∴ `"a"^2 - 10/"a"^2` = 3

∴ a⁴ – 10 = 3a²

∴ a⁴ –  3a² –  10 = 0

∴ (a² –  5)(a² + 2) = 0

∴ a² = 5 or a² = – 2

But a ∈ R

∴ a² ≠ – 2

∴ a² = 5

∴ a = `±sqrt(5)` 

When a = `sqrt(5)`, b = `(sqrt(10))/sqrt(5) = sqrt(2)`

When a = `-sqrt(5)`, b = `(sqrt(10))/(-5) = -sqrt(2)`

∴ `sqrt(3 + 2sqrt(10)"i") = ± (sqrt(5) + sqrt(2)"i")`