What is the square root of (8 + 6i)?  

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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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