4 solved out of 8 shown

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

The lower bound is easy: this is $\geq n$ and equality holds if and only if $p(z)=z^n$. The Chebyshev polynomial shows that the constant $1/2$ in this conjecture is best possible. Erdős originally conjectured this without the $o(1)$ term but Szabados observed that was too strng. Pommeranke has proved an upper bound of $O(n^2)$.

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].

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 forall $z\in\mathbb{C}$ 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 $\lvert z\rvert=1$.