Let $f$ be a polynomial such that $|f(z)| ≤ 1 − |z|^2 + |z|^{1000}$ for all $z ∈ C.$ Prove that $|f(0)| ≤ 0.2.$ 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 $|f(0)| ≤ 0.2.$
 A: 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: $1.1^{500} > 1+0.1\cdot 500 = 51$.
A: 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 $|z| \le t_0$.
