6 solved out of 13 shown

If $p(z)\in\mathbb{C}[z]$ is a monic polynomial of degree $n$ then is the length of the curve $\{ z\in \mathbb{C} : \lvert p(z)\rvert=1\}$ maximised when $p(z)=z^n-1$?

If $p(z)$ is a polynomial of degree $n$ such that $\{z : \lvert p(z)\rvert\leq 1\}$ is connected then is it true that
\[\max_{\substack{z\in\mathbb{C}\\ \lvert p(z)\rvert\leq 1}} \lvert p'(z)\rvert \leq (\tfrac{1}{2}+o(1))n^2?\]

The lower bound is easy: this is $\geq n$ and equality holds if and only if $p(z)=z^n$. The assumption that the set is connected is necessary, as witnessed for example by $p(z)=z^2+10z+1$.

The Chebyshev polynomials show that $n^2/2$ is best possible here. Erdős originally conjectured this without the $o(1)$ term but Szabados observed that was too strong. Pommerenke [Po59a] proved an upper bound of $\frac{e}{2}n^2$.

Eremenko and Lempert [ErLe94] have shown this is true, and in fact Chebyshev polynomials are the extreme examples.

Let $p(z)=\prod_{i=1}^n (z-z_i)$ for $\lvert z_i\rvert \leq 1$. Then the area of the set where \[A=\{ z: \lvert p(z)\rvert <1\}\]
is $>n^{-O(1)}$ (or perhaps even $>(\log n)^{-O(1)}$).

Conjectured by Erdős, Herzog, and Piranian [ErHePi58]. The lower bound $\mu(A) \gg n^{-4}$ follows from a result of Pommerenke [Po61]. The stronger lower bound $\gg (\log n)^{-O(1)}$ is still open.

Wagner [Wa88] proves, for $n\geq 3$, the existence of such polynomials with \[\mu(A) \ll_\epsilon (\log\log n)^{-1/2+\epsilon}\] for all $\epsilon>0$.

$100

Let $z_i$ be an infinite sequence of complex numbers such that $\lvert z_i\rvert=1$ for all $i\geq 1$, and for $n\geq 1$ let
\[p_n(z)=\prod_{i\leq n} (z-z_i).\]
Let $M_n=\max_{\lvert z\rvert=1}\lvert p_n(z)\rvert$. Is it true that $\limsup M_n=\infty$? Is it true that there exists $c>0$ such that for infinitely many $n$ we have $M_n > n^c$, or even that for all $n$
\[\sum_{k\leq n}M_k > n^{1+c}?\]

The weaker conjecture that $\limsup M_n=\infty$ was proved by Wagner, who show that there is some $c>0$ with $M_n>(\log n)^c$ infinitely often.

This was solved by Beck [Be91], who proved that there exists some $c>0$ such that \[\max_{n\leq N} M_n > N^c.\]

For every $\epsilon>0$ there is a polynomial $P(z)=\prod_{i=1}^n(z-\alpha_i)$ with $\lvert \alpha_i\rvert=1$ such that the measure of the set for which $\lvert P(z)\rvert<1$ is at most $\epsilon$.

Erdős says it would also be of interest to determine the dependence of $n$ on $\epsilon$ - perhaps $\epsilon>1/(\log n)^c$ for some $c>0$ is necessary.

Does there exist, for all large $n$, a polynomial $P$ of degree $n$, with coefficients $\pm 1$, such that
\[\sqrt{n} \ll \lvert P(z) \rvert \ll \sqrt{n}\]
for all $\lvert z\rvert =1$, with the implied constants independent of $z$ and $n$?

Originally a conjecture of Littlewood. The answer is yes (for all $n\geq 2$), proved by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST19].

See also [230].

Let $P(z)=\sum_{1\leq k\leq n}a_kz^k$ for some $a_k\in \mathbb{C}$ with $\lvert a_k\rvert=1$ for $1\leq k\leq n$. Does there exist a constant $c>0$ such that, for $n\geq 2$, we have
\[\max_{\lvert z\rvert=1}\lvert P(z)\rvert \geq (1+c)\sqrt{n}?\]

The lower bound of $\sqrt{n}$ is trivial from Parseval's theorem. The answer is no (contrary to Erdős' initial guess). Kahane [Ka80] constructed 'ultraflat' polynomials $P(z)=\sum a_kz^k$ with $\lvert a_k\rvert=1$ such that
\[P(z)=(1+o(1))\sqrt{n}\]
uniformly for all $z\in\mathbb{C}$ with $\lvert z\rvert=1$, where the $o(1)$ term $\to 0$ as $n\to \infty$.

For more details see the paper [BoBo09] of Bombieri and Bourgain and where Kahane's construction is improved to yield such a polynomial with \[P(z)=\sqrt{n}+O(n^{\frac{7}{18}}(\log n)^{O(1)})\] for all $z\in\mathbb{C}$ with $\lvert z\rvert=1$.

See also [228].

Let $f(k)$ be the minimum number of terms in $P(x)^2$, where $P\in \mathbb{Q}[x]$ ranges over all polynomials with exactly $k$ non-zero terms. Is it true that $f(k)\to\infty$ as $k\to \infty$?

First investigated by Rényi and Rédei [Re47]. Erdős [Er49b] proved that $f(k)<k^{1-c}$ for some $c>0$. The conjecture that $f(k)\to \infty$ is due to Erdős and Rényi.

This was solved by Schinzel [Sc87], who proved that \[f(k) > \frac{\log\log k}{\log 2}.\] In fact Schinzel proves lower bounds for the corresponding problem with $P(x)^n$ for any integer $n\geq 1$, where the coefficients of the polynomial can be from any field with zero or sufficiently large positive characteristic.

Schinzel and Zannier [ScZa09] have improved this to \[f(k) \gg \log k.\]

Is it true that all except $o(2^n)$ many polynomials of degree $n$ with $\pm 1$-valued coefficients have $(\frac{1}{2}+o(1))n$ many roots in $\{ z\in \mathbb{C} : \lvert z\rvert \leq 1\}$?

For any $t\in (0,1)$ let $t=\sum_{k=1}^\infty \epsilon_k(t)2^{-k}$ (where $\epsilon_k(t)\in \{0,1\}$). If $S_n(t)$ is the number of roots of $\sum_{1\leq k\leq n}\epsilon_k(t)z^k$ in $\lvert z\rvert \leq1$ then is it true that, for almost all $t\in (0,1)$, \[\lim_{n\to \infty}\frac{S_n(t)}{n}=\frac{1}{2}?\]

Erdős and Offord [EO56] showed that the number of real roots of a random degree $n$ polynomial with $\pm 1$ coefficients is $(\frac{2}{\pi}+o(1))\log n$.

See also [521].

For any $t\in (0,1)$ let $t=\sum_{k=1}^\infty \epsilon_k(t)2^{-k}$ (where $\epsilon_k(t)\in \{0,1\}$). Does there exist some constant $C>0$ such that, for almost all $t\in (0,1)$,
\[\max_{\lvert z\rvert=1}\left\lvert \sum_{k\leq n}\epsilon_k(t)z^k\right\rvert=(C+o(1))\sqrt{n\log n}?\]

Salem and Zygmund [SZ54] proved that $\sqrt{n\log n}$ is the right order of magnitude, but not an asymptotic.

For any $t\in (0,1)$ let $t=\sum_{k=1}^\infty \epsilon_k(t)2^{-k}$ (where $\epsilon_k(t)\in \{0,1\}$). What is the correct order of magnitude (for almost all $t\in(0,1)$) for
\[M_n(t)=\max_{x\in [0,1]}\left\lvert \sum_{k\leq n}\epsilon_k(t)x^k\right\rvert?\]

A problem of Salem and Zygmund [SZ54]. Chung showed that, for almost all $t$, there exist infinitely many $n$ such that
\[M_n(t) \ll \left(\frac{n}{\log\log n}\right)^{1/2}.\]
Erdős (unpublished) showed that for almost all $t$ and every $\epsilon>0$ we have $\lim_{n\to \infty}M_n(t)/n^{1/2-\epsilon}=\infty$.