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Let $f(z)=a_0+\cdots+a_n z^n$ be a complex polynomial with $a_j = \alpha_j + i \beta_j$ (with $\alpha_j, \beta_j \in \mathbb{R}$). Let $P(z)=\alpha_0 + \cdots + \alpha_n z^n$ and $Q(z)=\beta_0 + \cdots + \beta_n z^n$, so that $f(z)=P(z)+iQ(z)$. Assume that all the roots of $f$ are in the half-plane $\operatorname{Im} z > 0$.

How do you prove that all the roots of $P$ and $Q$ are real?

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@Phira, Of course! The other root.. –  Alex Dec 3 '11 at 18:13
    
I'm sorry, what does not fulfill the hypothesis? –  timofei Dec 3 '11 at 18:30
    
@timofei, I posed an example that did not satisfy the conclusion, but Phira was saying the example does not fulfill the hypothesis of the problem, so doesn't count. My guess (and that's all it is right now) is if you assume $P$ or $Q$ has a complex root, then as the coefficients of $P$ and $Q$ are real, the complex conjugate must also be a root, and this may force $f$ to have a root that's not in the upper half plane. –  Alex Dec 3 '11 at 18:45
    
Yes I've been trying something along this way. In fact if $z$ is a root of $P$ (for example) then $f(\bar{z})=-\overline{f(z)}$ so that there exists some $t \in [0,1]$ such that $\mathrm{Re} (f(tz+(1-t)\bar{z}))=0$... But maybe this is useless. –  timofei Dec 3 '11 at 19:04
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1 Answer

up vote 4 down vote accepted

The idea is that when you go along the real axis, $f(x)$ makes $n/2$ complete loops around $0$, so has to cross the horizontal axis $n$ times (and the same for the vertical axis).

To make this precise, integrate $f'(z)/f(z)$ along the loop $\gamma_R(t) = Re^{it}$ for $t \in [0; \pi]$, and $\gamma_R(t) = R\cos(t)$ for $t \in [\pi ; 2 \pi]$. For $R$ large enough, the loop contains all roots of $P$, so the residue theorem tells you that the integral is $2in \pi$. Meanwhile, as $R$ goes to infinity, the integral on the semi-circle part converges to $ni \pi$

Thus, $\int_\mathbb{R} f'(x)/f(x) dx = ni \pi$. This means that the argument of $f(x)$ moves by $n\pi$ while $x$ moves along the real axis.

Since $\lim_{x \to \infty} \textrm{Arg}(f(x)) = \textrm{Arg}(a_n)$, the case when $a_n$ is neither real nor purely imaginary is easy : $P(x)$ has to cross each axis at least $n$ times in order for the argument to move by $n \pi$. This means that $P$ and $Q$ have at least $n$ roots in $\mathbb{R}$. Since they are of degree $n$, they both have their $n$ roots in $\mathbb{R}$.

In the case where $a_n$ is real or purely imaginery, it means that one of the axis is crossed only $n-1$ times, but since the corresponding polynom is of degree $n-1$, it is again the case that both polynoms have all their roots in $\mathbb{R}$.

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Another way of seeing that first property is by writing $f(z) = a_n (z - z_1) \cdots (z - z_n)$ and considering how $arg(f(z))$ varies when $z$ moves along the real line. –  leonbloy Dec 3 '11 at 19:32
    
Very clear. Thank you. –  timofei Dec 4 '11 at 9:59
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