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I need some help with the following question:

Let $P$ be a complex polynomial such that $z \in \mathbb{R} \iff P(z) \in \mathbb{R}$. Show that $deg P = 1$

There's also a hint:

Define $P = u+iv$ and show that either $v_y \le 0$ or $v_y \ge 0$ on the real axis.

I've managed to prove that. Then using CR equations I deduced that $u$ is monotone on the real axis which proves that $deg P$ is odd. I'm not sure what to do next.

Any help will be greatly appreciated

Thank you

amirbd89
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    "I deduced that $u$ is monotone on the real axis" and hence, $P$ induces a bijection of $\mathbb{R}$ with itself. – Daniel Fischer Jun 20 '15 at 13:58
  • But $z^3$ is also a bijection of $\mathbb{R}$ with itself – amirbd89 Jun 20 '15 at 14:00
  • Yes, I was just about to say that. – Tim Raczkowski Jun 20 '15 at 14:00
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    Yes, but $P$ is supposed to have another property, which $z \mapsto z^3$ doesn't have. – Daniel Fischer Jun 20 '15 at 14:01
  • Yes, what I meant was that I don't know how to use the fact that $P$ is such a bijection to prove that $deg$ $P = 1$ – amirbd89 Jun 20 '15 at 14:03
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    Ah. What do you know about polynomials of degree $d$ [$d > 0$]? – Daniel Fischer Jun 20 '15 at 14:05
  • They have n roots and in this case all roots must be real, and since $P$ is a bijection of $\mathbb{R}$ with itself n = 1. Thank you! – amirbd89 Jun 20 '15 at 14:10
  • @DanielFischer Why not make that all into an answer, then? Since, you know, it is one. As for the question itself, when I was a student we liked to joke that "hint" means "this is something that could probably be used to get the answer, but need not be the most efficient, simple, beautiful, or even obvious way". Why a particular hint failed at that was then a matter of some speculation: the professor didn't know? he's messing with you? he doesn't want you to use the good way, because this way makes you use the current material? – zibadawa timmy Jun 21 '15 at 00:18

3 Answers3

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This is how I would prove that $P$ is of degree $1$:

Say $P = a_nz^n + \cdots + a_0$, let $s > 0$ be a real number large enough that on the circle $\gamma$ with radius $s$ centered at the origin, the $n$-th degree term of $P$ dominates all the other terms together. Now parametrize $\gamma$ by $z(t) = se^{it}$, for $0\leq t < 2\pi$.

As $t$ goes from $0$ to $2\pi$, we have that $z(t)$ goes around $\gamma$, and $P(z(t))$ goes around the origin $n$ times. That means it hits the positive real axis at least $n$ times, and the negative real axis at least $n$ times. But $z(t)$ only hits the real axis twice (once on the positive side, and once on the negative side), so we must have $n = 1$, otherwise we have a value for which $P(z)$ is real, but $z$ is not.

Arthur
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Let $n$ be the degree of $P$.

The hypothesis implies that all $n$ zeros of $P$ are real because $0$ is real. This implies that $P$ has real coefficients because $P(x)=a(x-a_1)\cdots(x-a_n)$ implies $a$ is real by taking $x$ any real number that is not a root of $P$.

Now consider the equations $P(x)=c$ with $c$ real. The solutions must all be real, for every $c$. This means that the line $y=c$ cuts the graph of $P$ exactly $n$ times, counted with multiplicity. But this cannot happen if $n>1$ because $P$ goes monotonically to $\pm \infty$ as $x \to \infty$.

lhf
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  • This is not a complex analysis answer... – lhf Jun 20 '15 at 14:09
  • Please see my answer - I would have preferred putting as a comment to yours, but ran out of space... Basically isn't your "monotonically" analytic (i.e., depends on CR)? – peter a g Jun 20 '15 at 19:26
  • I like this answer, and I think that using any complex analysis at all is overkill. I don't understand @peter's comment: no, this is freshman calculus! A little bit of detail is left to fill in here; in particular, one does not a contradiction this way when n = 2 (but no problem...). – Pete L. Clark Jun 20 '15 at 19:37
  • @PeteL.Clark - I'm obviously missing something then - why do we have that $P$ is monotone? Where do we have that $P'$ does not change sign? – peter a g Jun 20 '15 at 19:39
  • Oh, maybe this is the point: use calculus to show that for any nonconstant polynomial there is C such that if $x \geq C$ then $f$ is monotone and that if $x \geq -C$ then $f$ is monotone. ("Use" is a little strong: this is a familiar fact.) But then away from a bounded set, $f$ is at most two to one... – Pete L. Clark Jun 20 '15 at 19:40
  • @PeteL.Clark - OK! Agreed. In my defense, I wanted to use the hint... – peter a g Jun 20 '15 at 19:47
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    After @PeteL.Clark's comments, I see that you were probably not using the monotonicity of $P$, and modified my answer in consequence; by the way I liked your answer too (and had +1'd it)... – peter a g Jun 21 '15 at 00:16
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As you and lhf both noted, $P(z)$ has real coefficients; from the hypothesis the $v$ vanishes exactly on the real axis, we know that (possibly after replacing $P$ with $-P$) $v$ is strictly positive (say) on the upper half plane and strictly negative on the lower half plane. Therefore $v_y\ge0$ on the real axis. By $CR$, we know that $u_x \ge 0$ on the real axis, as you said. Hence, viewing $P$ as a polynomial function from the reals to the reals, we know that $P$ is nowhere decreasing. $P$ is not constant (otherwise the entire complex plane is mapped into the reals, contrary to hypothesis), so there is a $a\in {\mathbb R}$ such that $P'(a)$ is not zero. Let $c = P(a).$ Then, the polynomial $P(z) - c $ does not have a repeated root at $a$. Hence, if $P$ is not of degree one, there is a $b \not=a$ (which must also be real, by the premise of the question) such that $$c = P(b) = P(a).$$

But this is not possible (as $P$ is nowhere decreasing, and $P'(a)\not= 0.$)

Hence $P$ is of degree one.

peter a g
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  • Really like your solution. Would you mind explaining me simple assumptions you made at the beginning? – Meitar Dec 10 '15 at 16:12
  • Thank you! lhf explains nicely why $P$ is real - so are you asking about the sign of $v$? If so: the hypothesis is that $v$ vanishes exactly on the real axis, i.e., ${z\mid v(z) \not= 0 } = \mathbb C \setminus \mathbb R$. As $v$ is continuous, it must have constant sign when restricted to a component of $\mathbb C \setminus \mathbb R$. As $P$ is real, $P(\bar z)=\overline {P(z)} $, and we must have that $v$ takes opposite signs on each component. So, WLOG (by replacing $P$ with $-P$ if necessary), we can assume that $v$ is positive on the top component, and negative on the bottom. (cont.) – peter a g Dec 10 '15 at 16:43
  • (cont.) Hence, if one takes a vertical path going through the $x$-axis, $v$ must increase, so $v_y\ge 0$ along the $x$-axis. – peter a g Dec 10 '15 at 16:43
  • Just what I was thinking. Thank you very much. Highly precise. – Meitar Dec 11 '15 at 18:48
  • Why can $P$ not be constant along the real axis? – student Oct 17 '16 at 16:28
  • @student as I wrote in the answer above, if $P$ were constant along the real axis, $P$ would then be constant on the entire plane, contradicting the hypothesis that $P$ is real valued only on the real line. Does that answer your question? – peter a g Oct 17 '16 at 17:54
  • No, I'm sorry, I'm a little slow. Why does constant along the real axis imply constant on the entire plane? – student Oct 17 '16 at 18:37
  • @student Well, for instance, a non-zero poly $Q$ in 1 var with coeffs in a field has a finite number of roots - so if there exists a $c$ such that $P(z)= c$ holds for infinitely many $z$, the poly $Q = P -c $ vanishes on an infinite set, and hence is identically $0$, so $P = c$ identically. Other style of argument: say $f$ is an analytic function on a connected open set $\Omega$, and that $f$ is constant on a subset of $\Omega$ with an accumulation point $a\in\Omega$. Then $f$ is constant (I looked up a real reference - Rudin theorem 10.18 real and complex analysis). OK? – peter a g Oct 17 '16 at 19:40
  • @student I was sloppy in my first sentence in the preceding - " ... has a finite number of roots in the field " would have been better.... But do you follow? – peter a g Oct 17 '16 at 23:40
  • Thank you very much. Right. Your first argument didn't occur to me. I like it! Thank you. Now I just have one last question: If we only know $u_x$ and $v_y$ why is it enough to say something about $P'$? Don't we need to know that $v_x \ge 0$ on the real line also? – student Oct 18 '16 at 12:57
  • I found this and it seems to suggest that the derivative consists of $u_x$ and $v_x$. – student Oct 18 '16 at 12:58
  • @student In general, we have that $P = u + i v$, where $u$ and $v$ are real-valued. But, as $P$ is real-valued when restricted to the $x$-axis, one has here that $v$ is identically zero on the real axis. Therefore, for real $x$, one has that $P(x) = u(x,0)$, and that $P'(x) = u_x(x,0)$. – peter a g Oct 18 '16 at 14:03
  • Oh yes, right. Thank you very much for your explanation! – student Oct 18 '16 at 16:11