# Find the square root of the following complex numbers: – 8 – 6i - Mathematics and Statistics

Sum

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

#### 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 = a2 + b2i2 + 2abi
∴ – 8 – 6i = (a2 – b2) + 2abi       ...[∵ i2 = – 1]
Equating real and imaginary parts, we get
a2 – b2 = – 8 and 2ab = – 6
∴ a2 – b2 = – 8 and b = (-3)/"a"

∴ "a"^2 - (-3/"a")^2 = – 8

∴ "a"^2 - 9/"a"^2 = – 8

∴ a4 – 9 = – 8a2
∴ a4 + 8a2 – 9 = 0
∴ (a2 + 9)(a2 – 1) = 0
∴ a2 = – 9 or a2 = 1
But a ∈ R
∴ a2 ≠ – 9
∴ a2 = 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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