Question

Answer the following:

Find the square root of 6 + 8i

Answer 1

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

Squaring on both sides, we get

6 + 8i = a² + b²i² + 2abi

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

Equating real and imaginary parts, we get

a² − b² = 6 and 2ab = 8

∴ a² − b² = 6 and b = `4/"a"`

∴ `"a"^2 - (4/"a")^2` = 6

∴ `"a"^2 - 16/"a"^2` = 6

∴ a⁴ − 16 = 6a²

∴ a⁴ − 6a² − 16 = 0

∴ (a² − 8) (a² + 2) = 0

∴ a² = 8 or a² = −2

But a ∈ R 

∴ a² ≠ −2

∴ a² = 8

∴ a = `± 2sqrt(2)`

When a = `2sqrt(2)`, b = `4/(2sqrt(2)) = sqrt(2)`

When a = `-2sqrt(2)`, b = `4/(-2sqrt(2)) = -sqrt(2)`

∴ `sqrt(6 + 8"i") = ± (2 sqrt(2) + sqrt(2)"i") = ± sqrt(2)(2 + "i")`