I asked this question in math.stackexchange but I didn't have much luck. It might be more appropiate for this forum. Let $z\_1,z\_2,…,z\_n$ be i.i.d random points on the unit circle ($|z\_i|=1$) with uniform distribution on the unit circle. Consider the random polynomial $P(z)$ given by
$$
P(z)=\prod\_{i=1}^{n}(z−z\_i).
$$
Let $m$ be the maximum absolute value of $P(z)$ on the unit circle $m=\max\{|P(z)|:|z|=1\}$.


How can I estimate $m$? More specifically, I would like to prove that there exist $\alpha>0$ such that the following holds almost surely as $n\to\infty$
$$
m\geq \exp(\alpha\sqrt{n}).
$$
Or at least that for every $\epsilon>0$ there exists $n$ sufficiently large such that 
$$
\mathbb{P}(m\geq\exp(\alpha\sqrt{n}))>1-\epsilon
$$
for some $\alpha$ independent on $n$.


Any idea of what can be useful here?



Polynomials on the Unit Circle
Here is a more careful (EDIT: even more careful!) argument that gives an affirmative answer to the weaker version of the question (as stated in the edit to my previous post, I doubt that the stronger version is true).


The argument uses the following lemma, which ought to be known. If someone has a reference, please leave a comment.


Lemma:
Let $a\_1,\dots, a\_n$ be real numbers with each $a\_i\geq 1$, and let $X\_1,\dots,X\_n$ be independent random variables, each uniform on $\pm 1$. Let $I$ be an interval of length $2r$. Then $$Pr(a\_1X\_1+\cdots+a\_nX\_n\in I) \leq \frac{1+r}{\sqrt{\pi n/2}}.$$


Proof: 
Let $f(X)$ denote $a\_1X\_1+\cdots+a\_nX\_n$. 
In the Boolean lattice of all assignments of $\pm 1$ to the variables $X\_1,\dots,X\_n$, consider a random walk starting from the point where all $X\_i$'s are $-1$, and moving in $n$ steps to the point where they are all $+1$, in each step choosing uniformly and independently of the history a variable which is $-1$ and changing it to $+1$. 


What is the expectation of the number $N(I)$ of steps of this walk at which $f(X) \in I$? On one hand, $N(I)\leq 1+r$, since $f(X)$ increases by at least 2 in each step.


On the other hand, the probability that the walk passes through any given point in the Boolean lattice is at least $2^{-n}\sqrt{\pi n/2}$ (this probability is minimized at the middle level(s) of the lattice, and the claim follows by well-known estimates of the central binomial coefficient). Therefore 
$$EN(I) \geq \frac{\#\{X:f(X)\in I\}}{2^n}\cdot \sqrt{\pi n/2} = Pr(f(X)\in I) \cdot \sqrt{\pi n/2}.$$


It follows that $$Pr(f(X)\in I) \leq \frac{1+r}{\sqrt{\pi n/2}}. \qquad \square$$


As was explained in the earlier post, we can randomly choose $n$ pairs of opposite points $\{z\_i, -z\_i\}$, then find $z$ with $\left|P(z)P(-z)\right|=1$ given only this information, and finally fix the $z\_i$'s by $n$ independent coin flips. 


In order to apply the lemma, we want to have, before the coin flipping, $n/2$ pairs $z\_i, -z\_i$ making an angle of say at most $60$ degrees with $z, -z$, so that each of the $n/2$ corresponding coin flips determine the sign of a term of at least $\log 3$ in $\log\left|P(z)\right| - \log\left|P(-z)\right|$. Actually, after choosing the $n$ pairs $z\_i, -z\_i$, this a. a. s. holds for every $z$. The idea is to divide the circle into, say, 100 equally large sectors. With high probability, every pair of opposite sectors will contain at least $n/51$ pairs (as opposed to the expected number, $n/50$).


We now condition on the outcomes of the coin flips for the smaller terms (pairs $z\_i, -z\_i$ more or less orthogonal to $z$). The lemma above tells us that for any interval $I$ of length $4\alpha\sqrt{n}$, the probability that $\log\left|P(z)\right| - \log\left|P(-z)\right| \in I$ is at most $4\alpha/\sqrt{\pi} + O(1/\sqrt{n})$.


In particular, with probability at least $1-O(\alpha)$, the absolute value of $\log\left|P(z)\right| - \log\left|P(-z)\right|$ is at least $2\alpha\sqrt{n}$, and since $\log\left|P(z)\right|=-\log\left|P(-z)\right|$, $\max\left(\log\left|P(z)\right|, \log\left|P(-z)\right|\right)\geq \alpha\sqrt{n}$ as required.


