Question

Find the square root of the following complex numbers: – 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).