How find the maximum value of $|bc|$

Solution 1:

You can assume that $a,b,c$ are reals and $b,c\ge0$. For $c$ this is obvious. For $b$ you can get this substituting $\theta z$ for $z$, as you wrote in the question. Finally, you can replace the polynomial by $\frac{(az^2+bz+c)+(\bar{a}z^2+bz+c)}2$.

By the maximum principle it suffices to verify the condition on the circle.

At the point $e^{it}$ we have $$ 1 \ge |ae^{2it}+be^{it}+c|^2 = |a+be^{-it}+ce^{-2it}|^2 \\ = \big(a+b\cos t+c\cos 2t\big)^2 + \big(b\sin t+c\sin 2t\big)^2 \\ = \big(a+b\cos t+c\cos 2t\big)^2 + \big(b\sin t-c\sin 2t\big)^2 + 4bc \sin t \sin 2t \\ \ge 4bc \sin t \sin 2t. $$ The maximum of $\sin t \sin 2t$ is $\frac{16}{3\sqrt3}$ (at $t=\arcsin\frac{\sqrt2}{\sqrt3}$ where $\cos t=\frac1{\sqrt3}$, $\sin t=\frac{\sqrt2}{\sqrt3}$, $\cos 2t=-\frac13$ and $\sin 2t=\frac{2\sqrt2}{3}$). So, at this point we get the bound $bc\le \frac{3\sqrt3}{16}$.

Of course this is only a bound...

To preserve equality in the extreme case, we have to set $b=\frac{\sqrt3}{2\sqrt2}$, $c=\frac{3}{4\sqrt2}$, $a=-\frac1{4\sqrt2}$. Then $$ |ae^{2it}+be^{it}+c|^2 = |ae^{it}+b+ce^{-it}|^2 \\ = \big((a+c)\cos t+b\big)^2 + \big((a-c)\sin t\big)^2 \\ = (a-c)^2+b^2 +2(a+c)b\cos t+4ac\cos^2t \\ = \frac78+\frac{\sqrt3}4\cos t -\frac38\cos^2 t = 1 - \frac38\left(\cos t-\frac1{\sqrt3}\right)^2 \le 1. $$ Hence, $bc=\frac{3\sqrt3}{16}$ is possible.