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Sum

Find the square root of the following complex numbers: 7 + 24i

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#### Solution

Let `sqrt(7 + 24"i")` = a + bi, where a, b ∈ R

Squaring on both sides, we get

7 + 24i = (a + bi)^{2}

∴ 7 + 24i = a^{2} + b^{2}i^{2} + 2abi

∴ 7 + 24i = (a^{2} – b^{2}) + 2abi ...[∵ i^{2} = – 1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 7 and 2ab = 24

∴ a^{2} – b^{2} = 7 and b = `12/"a"`

∴ `"a"^2 - (12/"a")^2` = 7

∴ `"a"^2 - 144/"a"^2` = 7

∴ a^{4} – 144 = 7a^{2}

∴ a^{4} – 7a^{2} – 144 = 0

∴ (a^{2} – 16)(a^{2} + 9) = 0

∴ a^{2} = 16 or a^{2} = – 9

But a ∈ R

∴ a^{2} ≠ – 9

∴ a^{2} = 16

∴ a = ± 4

When a = 4, b = `12/4` = 3

When a = – 4, b = `12/(-4)` = – 3

∴ `sqrt(7 + 24"i")` = ± (4 + 3i).

Concept: Square Root of a Complex Number

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Chapter 3: Complex Numbers - Exercise 3.2 [Page 40]