Answer 1

Let `sqrt(-16 + 30"i")`= a + bi, where a, b ∈ R
Squaring on both sides, we get
– 16 + 30i = a² + b²i² + 2abi
∴ – 16 + 30i = (a² – b²) + 2abi       ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = – 16 and 2ab = 30

∴ a² – b² = – 16 and b = `15/"a"`

∴ `"a"^2 -(15/"a")^2 = - 16`

∴ `"a"^2 - 225/"a"^2` = – 16

∴ a⁴ –  225  –  16a²
∴ a⁴ + 16a² – 225 = 0
∴ (a² + 25)(a² – 9) = 0
∴ a² = – 25 or a² = 9
But a ∈ R                    
∴ a² ≠ – 25
∴ a² = 9
∴ a = ± 3
When a = 3, b = `15/3` = 5

When a = – 3, b = `15/(-3)` = – 5

∴ `sqrt(-16 + 30"i")` = ± (3 + 5i)