The algebraic form of an imaginary root is `z = x + i*y` , where x represents the real part and y represents the imaginary part. A polynomial can only have an even number of comeplex roots since if there exists a complex root, hence there also exists its complex conjugate.

The following example shows a quadratic equation with two complex conjugate roots, such that:

`x^2 - 4x + 13 = 0`

Using quadratic formula yields:

`x_(1,2) = (4+-sqrt(16 - 52))/2 => x_(1,2) = (4+-sqrt-36)/2`

`x_(1,2) = (4+-6*i)/2 => x_1 = 2 + 2i; x_2 = 2 - 3*i`

**Hence, as conclusion, if the problem allows, you may consider the real roots of a polynomial equation as complex, having the imaginary parts `y = 0` .**

Usually, a complex root comes in pairs. So, if an equation has a complex root, it has as rootÂ the conjugate of the complex root, too.

Let's say that z = a + bi is the complex root of an equation of n-th order. The equation will have as root the conjugate z' = a - bi, too.

For a quadratic, the complex roots eliminate the real roots.

The complex roots of the quadratic occur when the discriminant of the equation is negative.