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What does it tell me when I find that a polynomial has complex roots, except for the obvious fact that it crosses zero for these values?

To me it seems that the existance of complex roots must have some implication in the real domain as well, though I don't know what or why.

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  • $\begingroup$ real numbers must be extended to complex numbers $\endgroup$ – Jasser Sep 30 '14 at 12:54
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    $\begingroup$ I don't think it gives you that much. When a polynomial has real coefficients, a complex root $a+bi$ tells you that the complex conjugate $a-bi$ is also a root, and that $(x^2 + 2ax + a^2+b^2)$ is a factor. $\endgroup$ – Dan Uznanski Sep 30 '14 at 12:57
  • $\begingroup$ @user291957, I suppose you mean to say that the real numbers are not closed, therefore you must consider the full set of complex numbers? $\endgroup$ – Oebele Sep 30 '14 at 13:39
  • $\begingroup$ Yes you are right @Ale Strooisma. But don't know much about what it can represent. $\endgroup$ – Jasser Sep 30 '14 at 14:17
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    $\begingroup$ This question is a little unfocused. You could just as easily ask "what does a negative root signify?" Usually mathematical objects do not have significance on their own. It is only through interaction with other mathematical objects that they gain it -- the properties of one mathematical object may relate to or control the properties of another, and thus gain a significance and perhaps an interpretation. I've answered this question in the context of the interaction between polynomials and differential equations, though perhaps this wasn't the context you originally had in mind. $\endgroup$ – Antonio Vargas Sep 30 '14 at 14:50
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A complex root of a polynomial can have some significance itself when the roots of the polynomial have significance in general. One example that comes to mind where the roots of polynomials have a meaningful interpretation is in the field of dynamical systems.

Consider a matrix differential equation

$$ X' = AX, $$

where $A$ is a constant real $2\times2$ matrix and $X$ is a $2\times1$ vector of real functions. The system always has a zero solution $X = \mathbf 0$, and the eigenvalues of $A$, which are the roots of the polynomial equation

$$ \det(A - \lambda I) = 0, $$

determine how the flow of solutions to the differential equation behaves with respect to this zero solution. The expression $\det(A - \lambda I)$ is indeed a polynomial, called the characteristic polynomial of the matrix.

When both eigenvalues are real and positive the solutions flow away from the point $X = \mathbf 0$ as time increases. When the eigenvalues are both negative the solutions flow directly to $X = \mathbf 0$ instead. There are other cases too, but let's focus on these. Here are a couple plots to illustrate, the first with negative eigenvalues and the second with positive:

enter image description here

enter image description here

If there is one complex eigenvalue then there must actually be two complex eigenvalues since the coefficients of the characteristic polynomial are real, and the real parts of these eigenvalues are equal. The real part of these eigenvalues plays the same role as above: if it's positive the solutions tend toward the origin and if it's negative the solutions move away from it. What the imaginary part indicates is that there is some rotational quality to the flow of solutions. Instead of flowing directly to the origin the solution will rotate around it infinitely often as it approaches or moves away.

Here's a plot showing the flow of a system whose eigenvalues are $1\pm 2i$:

enter image description here

If the eigenvalues are purely imaginary (i.e. their real parts are zero) then the solutions no longer tend to or away from the origin. All that remains is their rotational component, and they travel in ellipses around the origin, never settling down, as in the following plot.

enter image description here

To summarize, the presence of complex roots of the characteristic polynomial tells us that the flow of solutions to the corresponding differential equation has a rotational quality which is not present when the roots are real. The roots being purely imaginary has the special significance in that the flow is purely rotational; the solutions no longer tend toward $\mathbf 0$ or $\mathbf \infty$.

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If a polynomial has complex roots, then in the real domain it will have less roots than would otherwise be "expected". When I say expected, we normally would think that a polynomial of degree $n$ would have $n$ roots, counting multiplicities, but this is only the case over an algebraically closed field (like $\mathbb C$, not $\mathbb R$). If you have $m$ complex roots then there will be $n-m$ real roots left (counting multiplicity).

For example if you have a quadratic polynomial in $x$ and $y$ with complex roots, then it is entirely in one half of the plane. Similarly a cubic in $x$ and $y$ with complex roots it will intersect $y=0$ once.

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  • $\begingroup$ As far as I know, there isn't much more to it in terms of implications in real space. There could maybe be something more to say in terms of the intersection of (polynomial) curves in projective space, but I'm guessing you weren't really asking about that, and it's really only a bit of an extension on the idea. $\endgroup$ – Dom Sep 30 '14 at 13:48
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    $\begingroup$ @AleStrooisma, further in this vein, if a root has a "small" imaginary part then the real graph of the polynomial will have a "dip" near the real part of the root. Indeed, if the imaginary part of the root (and its conjugate) tends to zero then the polynomial would have a double real root, and thus just touch the $x$-axis like a parabola would. So if the imaginary part of the root was small but nonzero then the graph will still look like a parabola near there but it won't intersect the axis -- it will dip down to it, but not all the way. $\endgroup$ – Antonio Vargas Sep 30 '14 at 17:46

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