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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

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  • $\begingroup$ Do you want all roots to come in complex conjugate pairs? In that case, it's not hard to verify that the only possibilities are the ones you call 'trivial', namely, those of the form $w p(x)$, where $w \in \Bbb C - \Bbb R$ and $p(x)$ is real. $\endgroup$
    – Travis Willse
    Jun 16, 2016 at 12:40

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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.

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  • $\begingroup$ Thanks very much. That makes sense. $\endgroup$
    – NumberCruncher
    Jun 22, 2016 at 12:00
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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.

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