Complex Conjugate roots with non real coefficients

Question

I understand that a polynomial with real coefficients must have complex conjugate roots (if complex roots exist)

Is it possible for a polynomial with non-real coefficients to have complex conjugate roots? If yes, could you give me an example of a quadratic equation with non-real coefficients that give complex conjugate solutions (except for the trivial cases such as I(x^2-4x+13)=0)

Thanks

Answer 1

For quadratic polynomials this is is not possible due to Vieta: if $x_1, \bar{x}_1$ are the roots the coefficients of the quadratic are $x_1+\bar{x}_1$ and $x_1 \bar{x}_1$ and these are both real. For higher order poylnomials it is possible, if not all roots come in complex-conjugate pairs, otherwise the complete poynomial is a product of real quadratics.

Answer 2

Given a complex number z, a polynomial of degree two having $z$ and $\bar z$ as roots is $(x-z)(x-\bar z) = x^2 -(z + \bar z)x +z\bar z$.

$z +\bar z$ is real and $z\bar z$ is real too.

Thus to obtain a polynomial of degree two with conjugate complex roots and complex coefficients you only have trivial examples like $3i(x^2+1)$ or the one you mentioned above.