Putnam

Algebra Difficulty 8.2 Shortlist Find the answer

Given real numbers $b_0, b_1, \dots, b_{2019}$ with $b_{2019} \neq 0$ , let $z_1,z_2,\dots,z_{2019}$ be the roots in the complex plane of the polynomial $P(z) = \sum_{k=0}^{2019} b_k z^k.$ Let $\mu = (|z_1| + \cdots + |z_{2019}|)/2019$ be the average of the distances from $z_1,z_2,\dots,z_{2019}$ to the origin. Determine the largest constant $M$ such that $\mu \geq M$ for all choices of $b_0,b_1,\dots, b_{2019}$ that satisfy $1 \leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019.$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The answer is $M = 2019^{-1/2019}$ . For any choices of $b_0,\ldots,b_{2019}$ as specified, AM-GM gives $\mu \geq |z_1\cdots z_{2019}|^{1/2019} = |b_0/b_{2019}|^{1/2019} \geq 2019^{-1/2019}.$ To see that this is best possible, consider $b_0,\ldots,b_{2019}$ given by $b_k = 2019^{k/2019}$ for all $k$ . Then $P(z/2019^{1/2019}) = \sum_{k=0}^{2019} z^k = \frac{z^{2020}-1}{z-1}$ has all of its roots on the unit circle. It follows that all of the roots of $P(z)$ have modulus $2019^{-1/2019}$ , and so $\mu = 2019^{-1/2019}$ in this case.

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