 4 solved out of 8 shown
If $p(z)$ 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 $n$ 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 $\max_{n\leq N} M_n > N^c.$ Additional thanks to: Winston Heap 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]. Additional thanks to: Mehtaab Sawhney 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$\lvert z\rvert=1$. Additional thanks to: Mehtaab Sawhney 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.