What is the square root of (8 + 6i)?
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justaguide
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The square root of the given complex number can be expressed as a complex number x + iy.
We have x + yi = sqrt (8 + 6i)
take the square of both the sides
=> (x + yi) ^2 = 8 + 6i
=> x^2 + y^2*i^2 + 2xyi = 8 + 6i
equate the real and complex coefficients
x^2 – y^2 = 8 and 2xy = 6
2xy = 6
=> xy = 3
=> x = 3/y
Substitute in x^2 – y^2 = 8
=> (3/y) ^2 – y^2 = 8
=> 9/y^2 – y^2 = 8
=> 9 – y^4 = 8y^2
=> y^4 + 8y^2 – 9 = 0
=> y^4 + 9y^2 – y^2 – 9 = 0
=> y^2(y^2 + 9) – 1(y^2 + 9) = 0
=> (y^2 – 1) (y^2 + 9) = 0
=> y^2 = 1 and y^2 = -9
we leave out y^2 = -9 as the coefficient y is real
y^2 = 1
=> y = 1, x = 3
and y = -1, x = -3
The square root of 8 + 6i = 3 + i ; -3 – i
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