Show that polynomial $X^{n+1}-aX^n+aX-1$ has only roots of module $1$

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 ? Thanks. • What is a polynom? Nov 9, 2018 at 23:01 • @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. Nov 9, 2018 at 23:03 2 Answers 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. • 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. Nov 10, 2018 at 13:55 • @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. Nov 10, 2018 at 14:11
• 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 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.