Question

Answer the following:

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)