Maximum value of a complex polynomial on the unit disk

Solution 1:

This was trickier than I first thought. Perhaps there is a simpler solution, but here is one approach: Let $\omega = \exp(2\pi i/n)$. Then $$ \sum_{j=0}^{n-1} {(\omega^k)}^j = \begin{cases} n, & k \in \{ 0, n \} \\ 0, & 1 \le k \le n-1. \end{cases} $$ (This is a well-known property of roots of unity, or follows from computing the geometric sum if you prefer.)

Hence \begin{align} \frac1n \sum_{j=0}^{n-1} f(\omega^j z) &= \frac1n \sum_{j=0}^{n-1} \sum_{k=0}^{n} a_k {(\omega^j)}^k z^k \\ &= \frac1n \sum_{k=0}^{n} a_k z^k \sum_{j=0}^{n-1} {(\omega^k)}^j = a_0 + a_nz^n. \end{align}

Consequently, we get \begin{align} |a_0| + |a_n| &= \max_{|z|=1} |a_0 + a_nz^n | \\ &= \max_{|z|=1} \Big| \frac1n \sum_{j=0}^{n-1} f(\omega^j z) \Big| \\ &\le \frac1n \sum_{j=0}^{n-1} \max_{|z|=1} \big| f(\omega^j z) \big| = \max_{|z|=1} |f(z)|. \end{align}