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Let $P = a_0 + a_1X + ... + a_nX^n$, such as $\forall k \in$ {$ 0, 2 , .. , n $}, $a_k \in \mathbb{C}$.
Prove that if all roots of P have strictly negative imaginary parts, then :
$R = \operatorname{Re}(a_0) + \operatorname{Re}(a_1)X + ... + \operatorname{Re}(a_n)X^n$ : splits completely over $\mathbb{R}$.
We're dealing with roots here, so I decided to use Vieta's formulas that relate $a_i$'s to the roots. I tried to prove first that all roots of R are real, but to no avail.
Any ideas?

Martin R
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AymaneMaaitat
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  • For polynomials with real coefficients, it's true that the complex roots come in conjugate pairs. That is, if $a+bi$ is a root, then we know $a-bi$ will also be a root. – Lockjaw Aug 24 '20 at 20:29

1 Answers1

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I found the following method in

where it is applied to Hurwitz polynomials. It can be adapted to your problem.

Let $P$ be a complex polynomial whose roots all lie in the lower halfplane. For $0 \le \lambda \le 1$ consider the polynomials $$ P_\lambda(z) = P(z) + \lambda \overline{P}(z) $$ where $\overline{P}$ is obtained from $P$ by replacing all coefficients by their complex conjugates.

We want to show that $P_1$ has only real roots. If that were not the case then $P_1$ must have a root in the upper halfplane.

The roots of $P_\lambda$ are continuous functions of $\lambda$, and all roots of $P_0$ are in the lower halfplane. It follows that there must be a $\lambda \in (0, 1)$ such that $P_\lambda$ has a root on the real axis. But for $x \in \Bbb R$ is $|P(x)| = |\overline{P}(x)|$ and therefore $|P_\lambda(x)| > 0$, so this cannot happen.

Martin R
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  • I don't understand what you mean by the assertion : "the roots of $P_\lambda$ continuous functions of $\lambda$" – AymaneMaaitat Aug 24 '20 at 21:56
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    @AymaneLazarus: The roots of a polynomial are continuous functions of the coefficients, see for example https://math.stackexchange.com/q/63196/42969. Roughly: Small changes in $\lambda$ cause small changes of the roots of $P_\lambda$. – Martin R Aug 24 '20 at 21:58
  • Very nice answer! I am pretty sure you meant $P_{\lambda}(x)\ne 0$ since you get those equal module numbers (conjugate) and $\lambda < 1$. – orangeskid Aug 25 '20 at 00:09
  • @orangeskid Yes, thanks! – Martin R Aug 25 '20 at 04:06