Question

Find the square root of the following complex number: −8 − 6i

Answer 1

Let `sqrt(-8 - 6"i")` = a + bi, where a, b ∈ R

Squaring on both sides, we get

−8 − 6i = (a + bi)²

∴ −8 − 6i = a² + b²i² + 2abi

∴ −8 − 6i = (a² − b²) + 2abi   ...[∵ i² = −1]

Equating real and imaginary parts, we get

a² − b² = − 8 and 2ab = − 6

∴ a² − b² = − 8 and b = `(-3)/"a"`

∴ `"a"^2 - (-3/"a")^2` = − 8

∴ `"a"^2 - 9/"a"^2` = − 8

∴ a⁴ − 9 = − 8a²

∴ a⁴ + 8a² − 9 = 0

∴ (a² + 9)(a² − 1) = 0

∴ a² = − 9 or a² = 1

But a ∈ R

∴ a² ≠ − 9

∴ a² = 1

∴ a = ± 1

When a = 1, b = `(-3)/1` = − 3

When a = − 1, b = `(-3)/(-1)` = 3

∴ `sqrt(-8 - 6"i")` = ± (1 − 3i)