Question

Find the square root of the following complex numbers: `3 + 2 sqrt(10)  "i"`

Answer 1

Let `sqrt(3 + 2sqrt(10)"i"` = a + bi, where a, b ∈ R
Squaring on both sides, we get
`3 + 2 sqrt(10)  "i"` = (a + bi)²
∴ `3 + 2 sqrt(10)  "i"` = a² + b²i² + 2abi
`3 + 2 sqrt(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 – 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)/(-sqrt(5)) = -sqrt(2)`

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