Question

Find the square root of 15 – 8i

Answer 1

Let `sqrt(15 - 8"i")`= a + bi, where a, b ∈ R
Squaring on both sides, we get
15 –  8i = a² + b²i² + 2abi
∴ 15 –  8i = (a² – b²) + 2abi       ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 15 and 2ab = – 8

∴ a² – b² = 15 and b = `(-4)/"a"`

∴ `"a"^2 - ((-4)/"a")^2` = 15

∴ `"a"^2 - 16/"a"^2` = 15

∴ a⁴ – 16 = 15a²
∴ a⁴ – 15a² – 16 = 0
∴ (a² –16)(a² + 1) = 0
∴ a² = 16 or a² = – 1
But a ∈ R
∴ a² ≠ – 1
∴ a² = 16
∴ a = ± 4
When a = 4, b = `(-4)/4` = – 1

When a = – 4, b = `(-4)/(-4)` = 1

∴ `sqrt(15 - 8"i")` = ± (4 – i).