Question

Find the square root of : 18i

Answer 1

Let `sqrt(18"i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
18i = a² + b²i² + 2abi
∴ 0 + 18i = a² – b² + 2abi         ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 0 and 2ab = 18

∴ a² – b² = 0 and b = `9/"a"`

∴ `"a"^2 - (9/"a")^2` = 0

∴ `"a"^2 - 81/"a"^2` = 0

∴ a⁴ – 81 = 0
∴  (a² – 9)(a² + 9) = 0
∴ a² = 9 or a² = – 9
But a ∈ R
∴ a² ≠ – 9
∴ a² = 9
∴ a = ± 3
When a = 3, b = `9/3` = 3

When a = – 3, b = `9/(-3)` = – 3

∴ `sqrt(18"i")` = ±(3 + 3i) = ±3(1 + i).