The Modulus of all the roots of a Polynomial are equal to $1$

Question

Suppose the real number $\lambda \in (0,1)$, and let $n$ be a positive integer. Prove that all roots of the polynomial $$f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k}$$ have modulus equal to $1.$

The Putnam problem 2014 B4 is similar: Show that for each positive integer $n,$ all the roots of the polynomial $\sum_{k=0}^n 2^{k(n-k)}x^k$ are real numbers.

Answer 1

Write the above polynomial in the form: $a+bx+cx^2$... and observe that if we put $f(x)=0$ it is same as putting $x^k f(1/x)=0$ {property of binomial coefficients and the power of lambda} conclude that if $x$ is a root then $1/x$ is also has the same modulus. Hence $|x|=1$

Answer 2

The following solution is taken from The Modulus of a Polynomial are the Same on AoPS. For other solutions see A family of polynomials whose zeros all lie on the unit circle on Math Overflow.

The idea is to derive a recursion formula for $$ f_n(z) = \sum_{k=0}^{n} \binom{n}{k}\lambda^{k(n-k)} z^k $$ and prove the statement by induction. $\lambda$ is a fixed real number in the interval $(0, 1)$.

$f_0(z) = 1$ and $f_1(z) = 1+z$ surely have only roots of modulus one.

Now assume that $n \ge 1$ and all roots of $f_n$ have modulus one. Using the recurrence formula for binomial coefficients $$ \binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1} $$ one gets $$ f_{n+1}(z) = f_n(\lambda z) + \lambda^n z f_n(\frac{z}{\lambda}) \, . $$ Now let $z^*$ be a zero of $f_{n+1}$. We have to show that $|z^*| = 1$. From the above recurrence formula we get that $$ f_n(\lambda z^*) = - \lambda^n z^* f_n(\frac{z^*}{\lambda}) $$ Note that $f_n(\frac{z^*}{\lambda})$ cannot be zero: Otherwise both $\lambda z^*$ and $\frac{z^*}{\lambda}$ would have modulus one, which is not possible.

Denoting the roots of $f_n$ with $z_1, \ldots z_n$ we have $f_n(z) = (z-z_1) \cdots (z-z_n)$ so that $$ z^* = - \frac{f_n(\lambda z^*)}{\lambda^n f_n(\frac{z^*}{\lambda})} = - \prod_{k=1}^n \frac{\lambda z^* - z_k}{z^* - \lambda z_k} $$ and therefore $$ \tag{*} |z^*| = \prod_{k=1}^n \left |\frac{\lambda z^* - z_k}{z^* - \lambda z_k} \right | \, . $$ An elementary calculation (using $|z_k|=1$ for all $k$) shows that $$ |\lambda z^* - z_k|^2 - |z^* - \lambda z_k|^2 = (1 - \lambda^2)(1 - |z^*|^2) \, . $$

It follows that if $|z^*| < 1$ then all factors on the right-hand side of $(*)$ are larger than one, and if $|z^*| > 1$ then all factors on the right-hand side of $(*)$ are less than one. So both cases lead to a contradiction, and the only possibility is that $|z^*| = 1$. This concludes the proof.