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Is there any necessary and sufficinet condition for a complex polynomial to have a real root?

A complex polynomial has a real root if and only if...?

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Sufficient, yes. Necessary, probably not. (If $p$ has a real root then so does $pq$ for any complex polynomial $q$.) –  Yemon Choi Jul 18 '11 at 11:06
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3 Answers 3

up vote 7 down vote accepted

Yes. If your polynomial is not yet real, replace $P$ by $P\bar P$ ($\bar P$ has complex conjugated coefficients). Therefore we may suppose that $P\in{\mathbb R}[X]$.Using the Euclid algorithm, you may find the g.c.d of $P$ and $P'$. Dividing $P$ by this g.c.d, your are left with the case where $P$ is real and has simple roots.

Now, you use te Euclid algorithm : $P_0=P$, $P_1=P'$ and $P_{k-1}=Q_kP_k-P_{k+1}$. The sequence $(P_k)_k$ ends with a constant polynomial. Take $a>0$ large enough that $P$ may not have a root in $[a,+\infty)$. Let $V(a)$ be the number of sign changes in the sequence $(P_k(a))_k$. Likewise, take $b<0$ such that $P$ has no root in $(-\infty,b)$ and compute $V(b)$.

Theorem : the number of real roots of $P$ equals $V(b)-V(a)$.

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This seems great, but is not a condition in terms of the coefficinets. Is there any condition obtained by the coefficients or by the discreminant of p? –  mosen Jul 18 '11 at 13:43
    
Denis: what is $Q_k$? –  Willie Wong Jul 18 '11 at 14:27
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@Willie: I think $Q_k$ is just the quotient when $P_{k-1}$ is divided by $P_k$. –  TonyK Jul 18 '11 at 16:29
    
@mosen. If $P$ is quadratic, this calculation gives you the discriminant. –  Denis Serre Jul 19 '11 at 9:15
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This is just a bit too long for a comment :)

Let $P$ be your polynomial and $x$ its real root. Obviously, $P(x)=0$ if and only if $$Q(x)\equiv (\mathrm{Re} P(x))^2+(\mathrm{Im} P(x))^2=0.$$ Now, $Q(x)$ is a polynomial with real coefficients, which reduces your question to finding criteria for a real-coefficients polynomial to have a real root, and these are discussed here.

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Why is $Q(z)$ a polynomial in $z$? –  Yemon Choi Jul 18 '11 at 12:58
    
Specifically, take $P(z)=z-1$ –  Yemon Choi Jul 18 '11 at 12:59
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@Yemon: $x$ is assumed to be real, and then my $Q(x)$ equals $P(x)\overline{P(x)}$ (cf. Denis' answer). –  mathphysicist Jul 18 '11 at 13:39
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@Yemon: In fact $Q(x)=P(x)\bar P(x)$ even for complex $x$ (sorry for a poor formulation in previous comment), but, and that's the main point, for real $X$ $P(x)\bar P(x)$ is clearly a polynomial in $x$. –  mathphysicist Jul 18 '11 at 14:38
    
@Yemon: of course I meant for real $x$. –  mathphysicist Jul 18 '11 at 15:00
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since, $P$ has a real root if and only if $p\bar{p}$ (Denis introduced this above) has. $p\bar{p}$ is a real polynomial and one can use Tarski's theorem.

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