Maximum of $|(z-a_1)\cdots(z-a_n)|$ on the unit circle

Solution 1:

The maximum value is $c=2$. As noted in the comments by Daniel Fischer, this is attained by $z^n-1$ whose maximum value on the circle is $2$.

I will now prove that this is optimal, and that for any polynomial $P(z)=(z-a_1)\cdots (z-a_n)$ with $|a_i|=1$, there exists $|w|=1$ such that $|P(w)|\geq 2$.

Proof: Assume the contrary. Then $|P(z)|<2$ on the unit circle. Letting $\eta=\inf\{|z|:\ |P(z)|=2\}$ compactness of the unit circle implies that $\eta>1$, and so there exists a circle of radius $1<r<\eta$ such that $|P(z)|<2$ on the boundary. Let $C=(-1)^{n} a_1\cdots a_n$ denote the constant coefficient of $P(z)$. Then since $|C|=1$, by Rouche's theorem applied to $H(z)=P(z)-2C$, the polynomial $H(z)$ will have no zeros inside the disk of radius $r$. This implies that every root $z_i$ of $H(z)$ must have $|z_i|\geq r$. However $C$, which has norm $1$, is the constant coefficient of $H(z)$, and equals the product of the roots of $H(z)$. This implies that $$1=|C|=\left|\prod_i z_i\right|\geq r^n>1,$$ which is impossible, and so the claim follows.

Solution 2:

In fact even more is true. Given any $n$ points $a_1,..a_n$ on the unit circle and weights $p_1,p_2...p_n\ge 0$ such that $p_1+p_2+...p_n=n$ one has $$\max_{|z|=1}|z-a_1|^{p_1}|z-a_2|^{p_2}...|z-a_n|^{p_n}\ge 2.$$ This was a conjecture from the paper by Vsevolod Lev and Sergey Konyagin from $2003$ and was proved in an extremely beautiful way by user "fedja" https://mathoverflow.net/questions/64099/the-maximum-of-a-polynomial-on-the-unit-circle

An alternative proof was given by Alexander Eremenko (see the same post), however, he used quite a powerful machinery (Dubinin theorem).