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

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