Lunar New Year Countdown Function

It is easy to show that $(x-1)f(x) \; = \; \sum_{k=1}^{13}x^k - 13$ If $|z| \le 1$ then $\left|\sum_{k=1}^{13}z^k\right| \;\le \; \sum_{k=1}^{13} |z|^k \; \le \; 13$ with equality precisely when $z=1$ , Thus $(z-1)f(z) \neq 0$ for $|z| \le 1$ except when $z=1$ . Since $f(1) = 91 \neq 0$ , we deduce that $f(z) \neq 0$ whenever $|z| \le 1$ , making the answer $\boxed{0}$ .