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Sum

Find the square root of the following complex numbers: – 8 – 6i

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#### Solution

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

Squaring on both sides, we get

– 8 – 6i = (a + bi)^{2}

∴ – 8 – 6i = a^{2} + b^{2}i^{2} + 2abi

∴ – 8 – 6i = (a^{2} – b^{2}) + 2abi ...[∵ i^{2} = – 1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = – 8 and 2ab = – 6

∴ a^{2} – b^{2} = – 8 and b = `(-3)/"a"`

∴ `"a"^2 - (-3/"a")^2` = – 8

∴ `"a"^2 - 9/"a"^2` = – 8

∴ a^{4} – 9 = – 8a^{2}

∴ a^{4} + 8a^{2} – 9 = 0

∴ (a^{2} + 9)(a^{2} – 1) = 0

∴ a^{2} = – 9 or a^{2} = 1

But a ∈ R

∴ a^{2} ≠ – 9

∴ a^{2} = 1

∴ a = ± 1

when a = 1, b = `(-3)/1` = – 3

when a = – 1, b = `(-3)/(-1)` = 3

∴ `sqrt(- 8 - 6"i")` = ± (1 – 3i).

Concept: Square Root of a Complex Number

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