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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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 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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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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