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Find the square root of the following complex number: 3+210i - Mathematics and Statistics

Sum

Find the square root of the following complex number: 

`3 + 2sqrt(10)"i"`

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Solution

Let `sqrt(3 + 2sqrt(10)"i")` = a + bi, where a, b ∈ R

Squaring on both sides, we get

`3 + 2sqrt(10)"i"` = a2 + b2i2 + 2abi

∴ `3 + 2sqrt(10)"i"` = (a2 – b2) + 2abi   ...[∵ i2 = – 1]

Equating real and imaginary parts, we get

a2 – b2 = 3 and 2ab = `2sqrt(10)`

∴ a2 – b2 = 3 and b = `(sqrt(10))/"a"`

∴ `"a"^2 - ((sqrt(10))/"a")^2` = 3

∴ `"a"^2 - 10/"a"^2` = 3

∴ a4 – 10 = 3a2

∴ a4 –  3a2 –  10 = 0

∴ (a2 –  5)(a2 + 2) = 0

∴ a2 = 5 or a2 = – 2

But a ∈ R

∴ a2 ≠ – 2

∴ a2 = 5

∴ a = `±sqrt(5)` 

When a = `sqrt(5)`, b = `(sqrt(10))/sqrt(5) = sqrt(2)`

When a = `-sqrt(5)`, b = `(sqrt(10))/(-5) = -sqrt(2)`

∴ `sqrt(3 + 2sqrt(10)"i") = ± (sqrt(5) + sqrt(2)"i")`

Concept: Square Root of a Complex Number
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