I had posted the following problem on [stack exchange](https://math.stackexchange.com/questions/2722893/the-modulus-of-a-polynomial-are-the-same-is-1) before.



> 
> Suppose $\lambda$ is a real number in $\left( 0,1\right)$, and let $n$ be a positive integer. Prove that all the 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 $1$.
> 
> 
> 


I do not seem to know how to do it, to show if the roots of the polynomial have modulus one. [Putnam 2014 B4](https://math.stackexchange.com/questions/1054826/show-all-roots-of-sum-k-0n-2kn-k-xk-are-real-december-6-2014-putna?rq=1) 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.
This problem is very similar to the Putnam problem.



A family of polynomials whose zeros all lie on the unit circle
This is a special case of the Lee-Yang theorem. Let $G$ be a finite graph with vertex set $V$ and edge set $E$. Let $\beta \in (0,1)$ be a real number. Let $\sigma$ denote a ``spin function" $\sigma: V \to \{ -1, 1\}$. Put $m(\sigma)$ to be the number of vertices with positive spin, and $d(\sigma)$ to be the number of edges with vertices of opposite spin. Form the polynomial 
$$ 
Z\_\beta(x) = \sum\_{\sigma} \beta^{d(\sigma)} x^{m(\sigma)}, 
$$ 
where the sum is over all possible choices of the spins $\sigma$. The Lee-Yang theorem then states that all the zeros of $Z\_{\beta}(x)$ lie on the unit circle. Amazing! 


Apply the Lee-Yang theorem to the complete graph on $n$ vertices. The polynomial $Z\_{\beta}(x)$ in this case is exactly $\sum\_{k=0}^{n} \binom{n}{k} \beta^{k(n-k)} x^k$, and so this polynomial has all its roots on the unit circle. 


For a discussion of the Lee-Yang theorem see these [notes](https://math.berkeley.edu/~nikhil/courses/270/lec15.pdf) from a course by Nikhil Srivastava. The original Lee-Yang paper can be found [here](https://journals.aps.org/pr/abstract/10.1103/PhysRev.87.410) (behind a paywall). For more work on the Lee-Yang theorem (among other things), see [Borcea and Branden](https://arxiv.org/pdf/0809.3087.pdf). In particular, Theorem 8.4 there states the following more general form of the Lee-Yang theorem: Let $A=(a\_{ij})$ be a Hermitian $n\times n$ matrix with all entries inside the closed unit disc. Then the polynomial 
$$ 
f(z) = \sum\_{S\subset [n]} z^{|S|} \prod\_{\substack{i\in S \\ j\not\in S}} a\_{ij} 
$$ 
has all its zeros on the unit circle. (Here $S$ runs over all subsets of $[n]=\{1, \ldots, n\}$.)


