Let $f$ be a polynomial such that $|f(z)| ≤ 1 − |z|^2 + |z|^{1000}$ for all $z ∈ C.$ Prove that $|f(0)| ≤ 0.2.$

Question

I am working on an old qualifying exam problem and I can't seem to really get anywhere. I would love some help. Thank you.

Let $f$ be a polynomial such that $|f(z)| ≤ 1 − |z|^2 + |z|^{1000}$ for all $z ∈ C.$ Prove that $|f(0)| ≤ 0.2.$

Answer 1

The minimum value of $\phi(t) = 1 - t^2 + t^{1000}$ on the set $[0,\infty)$ is found easily enough: since $\phi'(t) = -2t + 1000 t^{999} = -2t(1-500t^{998})$ the minimum occurs at $t_0 = \sqrt[998]{1/500}$. Thus if $|z| = t_0$, we have $$|f(z)| \le 1 - t_0^2 + t_0^{1000}$$ which is (computation omitted thanks to Wolfram Alpha) less than $0.2$. The maximum modulus principle now implies that $|f(z)| \le 0.2$ for all $|z| \le t_0$.

Answer 2

This is a strange question to put on a complex analysis qual, since numerical estimates overshadow the complex analysis material (the maximal principle, as in the answer by Umberto P.).

We need a value of $|z|$ such that $1-|z|^2+|z|^{1000}$ can be bounded by $1/5$. I'll take $$|z| = \sqrt{\frac{10}{11}}$$ so that the desired inequality becomes $$\frac{1}{11} + \left(\frac{10}{11}\right)^{500} \le \frac15$$ It suffices to show that $$\left(\frac{10}{11}\right)^{500} \le \frac1{10}$$ which follows from Bernoulli's inequality: $1.1^{500} > 1+0.1\cdot 500 = 51$.