Sum

Find the square root of: `2 + 2 sqrt(3)"i"`

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#### Solution

Let `sqrt(2 + 2sqrt(3)"i")` = a + bi, where a, b ∈ R.

Squaring on both sides, we get

`2 + 2 sqrt(3)"i"` = a^{2} + b^{2}i^{2} + 2abi

∴ `2 + 2 sqrt(3)"i"` = a^{2} – b^{2} + 2abi ...[∵ i^{2} = – 1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 2 and 2ab = `2sqrt(3)`

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

∴ `"a"^2 - (sqrt(3)/"a")^2` = 2

∴ `"a"^2 - 3/"a"^2` = 2

∴ a^{4} – 3 = 2a^{2}

∴ a^{4} – 2a^{2} – 3 = 0

∴ (a^{2} – 3)(a^{2} + 1) = 0

∴ a^{2} = 3 or a^{2} = – 1

But a ∈ R

∴ a^{2} ≠ – 1

∴ a^{2} = 3

∴ a = ± `sqrt(3)`

When a = `sqrt(3), "b" = sqrt(3)/sqrt(3)` = 1

When a = `-sqrt(3), "b" = sqrt(3)/-sqrt(3)` = – 1

∴ `sqrt(2 + 2sqrt(3)"i") = ± (sqrt(3) + "i")`.

Concept: Square Root of a Complex Number

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