Question

Find the square root of the following complex numbers: 7 + 24i

Answer 1

Let `sqrt(7 + 24"i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 + 24i = (a + bi)²
∴ 7 + 24i = a² + b²i² + 2abi
∴ 7 + 24i = (a² – b²) + 2abi         ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 7 and 2ab = 24

∴ a² – b² = 7 and b = `12/"a"`

∴ `"a"^2 - (12/"a")^2` = 7

∴ `"a"^2 - 144/"a"^2` = 7

∴ a⁴ – 144 = 7a²
∴ a⁴ – 7a² – 144 = 0
∴ (a² – 16)(a² + 9) = 0
∴ a² = 16 or a² = – 9
But a ∈ R
∴ a² ≠ – 9
∴ a² = 16
∴ a = ± 4
When a = 4, b = `12/4` = 3

When a = – 4, b = `12/(-4)` = – 3

∴ `sqrt(7 + 24"i")` = ± (4 + 3i).