I gave an exercise to a student the other day, thinking I had a simple solution for it, but it seems that my solution was just bullshit (at least, not conclusive). Well, at least, I warned her I wasn't sure of my solution :-)

Here it is : $a\in[-1,1]$, show that all roots of polynomial $P=X^{n+1}-aX^n+aX-1$ have modulus $1$ (I try different ways of expressing the problem, tell me which one is better english...).

It is easy to see that

  1. roots of $P$ are $1$, $-1$ for some special cases, and complex not real roots $z_i$ such that $\overline{z_i}$ is also a root,
  2. as $P$ is kind of reciprocal, $z$ root implies $\frac1z$ is also a root$.

All the other things I tried are not conclusive. For example, $a\in[-1,1]$ may be translated as $a=\cos\theta$ for some $\theta\in\mathbb R$, but I don't see how to use this. I tried some rewriting, but nothing seems to work. I tried to work on the modulus of a root, tried the relations roots-coefficients...

I'm quite stuck here. Could you help me please ?


  • $\begingroup$ What is a polynom? $\endgroup$ Commented Nov 9, 2018 at 23:01
  • 1
    $\begingroup$ @WilliamElliot polynomial in some languages, I fixed the title. Note that, with the ial ending, polynomial really should be an adjective rather than a noun. $\endgroup$
    – Will Jagy
    Commented Nov 9, 2018 at 23:03

2 Answers 2


Let $C = \{ z \in \mathbb{C} : |z| = 1 \}$ and $D = \{ z \in \mathbb{C} : |z| < 1 \}$ be the unit circle and open unit disk. We will assume $a \ne \pm 1$ as their cases are trivial.

Your statement is true. We are going to prove following generalization:

For $\alpha_1, \alpha_2, \ldots, \alpha_m \in D$, define $f(z) = \prod\limits_{k=1}^m(z - \alpha_k)$ and $g(z) = \prod\limits_{k=1}^m (1-\bar{\alpha}_k z)$.
The polynomial $f(z) - g(z)$ has all its roots belong to $C$.

Consider their ratio $h(z) = \frac{f(z)}{g(z)}$. Since all $|\alpha_k| < 1$, $g(z)$ is never zero over $C$ and $h(z)$ is well defined there.

For $z \in C$, we have $$|h(z)| = \prod\limits_{k=1}^m \left|\frac{z-\alpha_k}{1-\bar{\alpha}_k z}\right| = \prod\limits_{k=1}^m \left|\frac{z-\alpha_k}{(\bar{z} - \bar{\alpha}_k) z}\right| = 1$$ The ratio $h(z)$ maps $C$ to $C$.

For each factor $\frac{z-\alpha_k}{1-\bar{\alpha}_k z}$, when $z$ move long $C$ once, the factor move along $C$ also once. This implies their product $h(z)$ move along $C$ exactly $m$ times. As a result, we can find $m$ distinct $\theta_1, \ldots, \theta_m \in [ 0, 2\pi )$ such that $$h(e^{i\theta}) = 1 \iff f(e^{i\theta}) - g(e^{i\theta}) = 0$$ Polynomial $f(z) - g(z)$ has at least $m$ distinct roots over $C$. Since degree of $f(z) - g(z)$ is $m$, counting multiplicity, it has exactly $m$ roots in $\mathbb{C}$. This means above $m$ roots on $C$ is all the roots of $f(z) - g(z)$ and all of them are simple.

On the special case $m = n + 1$ and $(\alpha_1,\alpha_2,\ldots,\alpha_m) = (a,0,\ldots,0)$ where $a \in (-1,1)$. Polynomial $f(z) - g(z)$ reduces to

$$z^n(z - a) - (1-az) = z^{n+1} - a z^n + az - 1 = P(z)$$

and your statement follows.

IMHO, this is a good chance to introduce the concept of winding number to the students. If they are not ready for that. A standalone proof for the original statement (again $a \ne \pm 1$) goes like this.

When $a \in (-1,1)$, parameterize $C$ by $[0,2\pi) \ni \theta \mapsto z \in C$. We have

$$P(z) = z^{n+1} - az^n + az - 1 = 2ie^{i\frac{(n+1)\theta}{2}} \left[\sin\frac{(n+1)\theta}{2} - a\sin\frac{(n-1)\theta}{2}\right]$$ Let's call what's inside the square bracket as $I(\theta)$.

When $a$ is real, $I(\theta)$ is clearly real and $\theta = 0$ is a root of it.

Let $\theta_k = \frac{(2k+1)\pi}{n+1}$ for $k = 0,\ldots,n$. When $a \in (-1,1)$, it is easy to see $I(\theta_k)$ is positive for even $k$ and negative for odd $k$. This means $I(\theta)$ has $n$ more roots. One root from each interval $(\theta_{k-1},\theta_k)$ for $k = 1,\ldots, n $.

As a result, $I(\theta)$ has at least $n+1$ roots over $[0,2\pi)$. This is equivalent to $P(z)$ has at least $n+1$ roots on $C$. Once again, since $P(z)$ has degree $n+1$, these are all the roots it has.

  • $\begingroup$ If I could understand why $\left|\frac{z-\alpha_i}{1-\alpha_iz}\right|=1$, I would have a much simpler proof. But I don't see how this is true. $\endgroup$ Commented Nov 10, 2018 at 13:55
  • $\begingroup$ @NicolasFRANCOIS I make a mistake. It should be $\left|\frac{z - \alpha}{1 - \bar{\alpha} z}\right|$. notice there is a complex conjugate of $\alpha$ in denominator. $\endgroup$ Commented Nov 10, 2018 at 14:11
  • $\begingroup$ OK. I think I'll stick to the second demonstration for the moment. Thanks a lot for your help. I got the factorization, but couldn't see how to use it. A simple function study would have done the trick ! Silly me :-P $\endgroup$ Commented Nov 10, 2018 at 14:16

The equation can be rewritten as $$ B(x)=x^n\frac{x-a}{1-ax}=1. $$ The function $B$ is analytical in the unit disc for $a\in(-1,1)$ (the cases $a=\pm 1$ are simple), and $|B(x)|=1$ on the unit circle. Now apply the maximum modulus principle to $B(x)$ and $B(1/x)$ to ensure that the equality is only possible on the unit circle.


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