51. Let complex numbers $a, b, c$ satisfy that for any complex number $z$ with modulus not exceeding 1, $\left|a z^{2}+b z+c\right| \leqslant 1$. Find the maximum value of $|bc|$. (2010 China Mathematical Olympiad Problem)
$$\begin{array}{l}
51 \text { Let } f(z)=a z^{2}+b z+c, g(z)=z^{2} f(z)=a+b z^{+1}+c z^{-2}, h(-z)= \\
\mathrm{e}^{i \alpha} g\left(\mathrm{e}^{i \beta} \bar{z}\right)=c^{\prime} z^{-2}+b^{\prime} z^{-1}+a^{\prime}
\end{array}$$

Take appropriate real numbers $\alpha, \beta$, such that $c^{\prime}, b^{\prime} \geqslant 0$, for $r \leqslant 1$, we have
$$\frac{1}{r^{2}} \geqslant\left|h\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right| \geqslant\left|\cdot \operatorname{Im} h\left(r \mathrm{e}^{\mathrm{i} \theta}\right)-\right|=\left|r^{-2} c^{\prime} \sin 2 \theta+r^{-\mathrm{r}} b^{\prime} \sin \theta+\operatorname{Im} a^{\prime}\right|$$

Assume $\operatorname{Im} \boldsymbol{a}^{\prime} \geqslant 0$, otherwise we can make an appropriate substitution $\boldsymbol{\theta} \rightarrow-\boldsymbol{\theta}$, so for any $\theta(0<$ $\left.\theta<\frac{\pi}{2}\right)$, we have
$$\frac{1}{r^{2}} \geqslant r^{-2} c^{\prime} \sin 2 \theta+r^{-1} b^{\prime} \sin \theta \geqslant 2 r^{-\frac{3}{2}} \sqrt{b^{\prime} c^{\prime} \sin 2 \theta \sin \theta} \Rightarrow$$
$|b c|=b^{\prime} c^{\prime} \leqslant \frac{1}{4 r \sin 2 \theta \sin \theta}\left(\right.$ for any $\left.r \leqslant 1,0<\theta<\frac{\pi}{2}\right) \Rightarrow$
$$|b c| \leqslant \min _{r \leqslant 1,0<\theta<\frac{\pi}{2}} \frac{1}{4 r \sin 2 \theta \sin \theta}=\min _{0<\theta<\frac{\pi}{2}} \frac{1}{4 \sin 2 \theta \sin \theta}=\frac{1}{4 \max _{0<\theta<\frac{\pi}{2}} \sin 2 \theta \sin \theta}=\frac{3 \sqrt{3}}{16}$$
An example where $|b c|=\frac{3 \sqrt{3}}{16}$: $f(z)=\frac{\sqrt{2}}{8} z^{2}-\frac{\sqrt{6}}{4} z-\frac{3 \sqrt{2}}{8}$.
For $z=r e^{i \theta}(r \leqslant 1)$, we have
$$\begin{aligned}
\left|f\left(\mathrm{e}^{i \theta}\right)\right|^{2}= & \frac{1}{32}\left[\left(r^{2} \cos 2 \theta-2 \sqrt{3} r \cos \theta-3\right)^{2}+\left(r^{2} \sin 2 \theta-2 \sqrt{3} r \sin \theta\right)^{2}\right]= \\
& \frac{1}{32}\left[2 r^{4}+12 r^{2}+18-\left(2 \sqrt{3} r \cos \theta+r^{2}-3\right)^{2}\right] \leqslant \\
& \frac{1}{32}\left(2 r^{4}+12 r^{2}+18\right) \leqslant
\end{aligned}$$