Let $a,b,c,z$ be complex numbers such that for all $|z| \le 1$ we have $|az^2+bz+c| \le 1$. How to prove that $|bc| \le \frac{3 \sqrt{3} }{16}$?
Asked
Active
Viewed 224 times
8
-
after a wrong answer, it seems that $|b c| \leq \frac{1}{4}$, could you give me an example where $| b c| > \frac{1}{4}$? – Sep 16 '17 at 13:33
-
Yes, it is duplicate as Martin R says. (however a different answer would be very interesting). – Ataulfo Sep 16 '17 at 14:47