3
$\begingroup$

I've seen a few conflicting pieces of information online.

So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?

At very least could you give me a counterexample? A cubic with no real roots.

$\endgroup$
3
  • 5
    $\begingroup$ It's more clear to say "at least one real root" or "a real root". There are real cubics that have three real roots (there are none that have exactly two though). $\endgroup$ – quid♦ Mar 10 '19 at 14:47
  • 6
    $\begingroup$ @quid: If you count with multiplicity, that's true. If you don't, then you can have $f(x) = (x-1)(x-2)^2$. $\endgroup$ – Kevin Mar 10 '19 at 17:22
  • 1
    $\begingroup$ Indeed @Kevin I should have made that explicity. $\endgroup$ – quid♦ Mar 10 '19 at 19:15
29
$\begingroup$

One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.

$\endgroup$
17
$\begingroup$

As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.

A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.

$\endgroup$
2
  • $\begingroup$ @Glen_b It is reasonable to say has one X to mean there exists an X, though it would be more clear to say there is at least one X. $\endgroup$ – jgon Mar 12 '19 at 20:21
  • 1
    $\begingroup$ @Glen_b I have edited to remove the ambiguity. $\endgroup$ – Servaes Mar 12 '19 at 20:22
1
$\begingroup$

Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $a(x-u)(x-v)(x-w)$ where $u, v, w$ are the roots (they don't need to be distict) and $a$ is the leading coefficient. This lets us form polynomials with only complex roots such as $(x-i)^3$.

However, if all the original coefficients of the polynomial are real, and $c$ is a complex root, then its conjugate $\bar{c}$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.

This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.