Sum

Find the square root of: – 16 + 30i

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#### Solution

Let `sqrt(-16 + 30"i")`= a + bi, where a, b ∈ R

Squaring on both sides, we get

– 16 + 30i = a^{2} + b^{2}i^{2} + 2abi

∴ – 16 + 30i = (a^{2} – b^{2}) + 2abi ...[∵ i^{2} = – 1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = – 16 and 2ab = 30

∴ a^{2} – b^{2} = – 16 and b = `15/"a"`

∴ `"a"^2 -(15/"a")^2 = - 16`

∴ `"a"^2 - 225/"a"^2` = – 16

∴ a^{4} – 225 – 16a^{2}

∴ a^{4} + 16a^{2} – 225 = 0

∴ (a^{2} + 25)(a^{2} – 9) = 0

∴ a^{2} = – 25 or a^{2} = 9

But a ∈ R

∴ a^{2} ≠ – 25

∴ a^{2} = 9

∴ a = ± 3

When a = 3, b = `15/3` = 5

When a = – 3, b = `15/(-3)` = – 5

∴ `sqrt(-16 + 30"i")` = ± (3 + 5i)

Concept: Square Root of a Complex Number

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