Question

Find the square root of: 3 – 4i

Answer 1

Let `sqrt(3 - 4"i")`= a + bi, where a, b ∈ R
Squaring on both sides, we get
3 – 4i = a² + b²i² + 2abi
∴ 3 – 4i = a² – b² + 2abi           ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 3 and 2ab = – 4

∴ a² – b² = 3 and b = `(-2)/"a"`

∴ `"a"^2 - (-2/"a")^2` = 3

∴ `"a"^2 - 4/"a"^2` = 3

∴ a⁴ – 4 = 3a²
∴ a⁴ – 3a² – 4 = 0
∴ (a² – 4)(a² + 1) = 0
∴ a² = 4 or a² = – 1
But a ∈ R
∴ a² ≠ – 1
∴ a²  = 4

∴ a = ± 2
When a = 2, b = `(-2)/2` = – 1

When a = – 2, b = `(-2)/(-2)` = 1

∴ `sqrt(3 - 4"i")` = ± (2 –  i).