设复数 $a、b、c$ 满足：对任意模不超过 $1$ 的复数 $z$，都有 $\left|a z^{2}+b z+c\right| \leqslant 1$．求 $|bc|$ 的最大值．
【难度】
【出处】
2010第25届CMO试题
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【答案】
略
【解析】
令 $f(z)=a z^{2}+b z+c,g(z)=z^{-2} f(z)=a+b z^{-1}+c z^{-2},h(z)=e^{i \alpha} g\left(e^{i \beta} z\right)=c^{\prime} z^{-2}+b^{\prime} z^{-1}+a^{\prime}$
取适当的实数 $\alpha, \beta$，使得 $c^{\prime}, b^{\prime} \geqslant 0$，对 $r \leqslant 1$，有 $\dfrac{1}{r^{2}} \geqslant\left|h\left(r \mathbf{e}^{\mathrm{i} \theta}\right)\right| \geqslant\left|\operatorname{Im} h\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right|$
不妨设 $\operatorname{Im} a^{\prime} \geqslant 0$，否则可作变换 $\theta \rightarrow-\theta$
这样对任意的 $\theta\left(0<\theta<\dfrac{\pi}{2}\right)$，有
$\dfrac{1}{r^{2}} \geqslant r^{-2} c^{\prime} \sin 2 \theta+r^{-1} b^{\prime} \sin \theta \geqslant 2 r^{-\frac{3}{2}} \sqrt{b^{\prime} c^{\prime} \sin 2 \theta \cdot \sin \theta}\Rightarrow|b c|=b^{\prime} c^{\prime} \leqslant \dfrac{1}{4 r \sin 2 \theta \cdot \sin \theta}$
（对任意 $r \leqslant 1, \theta \in\left(0, \dfrac{\pi}{2}\right)$）
$\Rightarrow|b c| \leqslant \min\limits _{r \leqslant 1, \theta \in\left(0 , \frac{\pi}{2}\right)} \dfrac{1}{4 r \sin 2 \theta \cdot \sin \theta}=\min\limits _{\theta \in(0 ,\frac{\pi}{2})}\dfrac{1}{4 \sin 2 \theta \cdot \sin \theta}=\dfrac{1}{4 \max \limits_{\theta \in\left(0,\frac{\pi}{2}\right)} \sin 2 \theta \cdot \sin \theta}=\dfrac{3 \sqrt{3}}{16}\cdot|b c|=\dfrac{3 \sqrt{3}}{16}$
的例子：$f(z)=\dfrac{\sqrt{2}}{8} z^{2}-\dfrac{\sqrt{6}}{4} z-\dfrac{3 \sqrt{2}}{8}$ 对于 $z=r \mathrm{e}^{\mathrm{i} \theta}(r \leqslant 1)$，有
$\left|f\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right|^{2}=\dfrac{1}{32}\left[\left(r^{2} \cos 2 \theta-2 \sqrt{3} r \cos \theta-3\right)^{2}+\right.\left(r^{2} \sin 2 \theta-2 \sqrt{3} r \sin \theta\right)^{2} ]\\=\dfrac{1}{32}\left[2 x^{4}+12 r^{2}+18-\left(2 \sqrt{3} r \cos \theta+r^{2}-3\right)^{2}\right]\leqslant \dfrac{1}{32}\left(2 r^{4}+12 r^{2}+18\right) \leqslant 1$
取适当的实数 $\alpha, \beta$，使得 $c^{\prime}, b^{\prime} \geqslant 0$，对 $r \leqslant 1$，有 $\dfrac{1}{r^{2}} \geqslant\left|h\left(r \mathbf{e}^{\mathrm{i} \theta}\right)\right| \geqslant\left|\operatorname{Im} h\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right|$
不妨设 $\operatorname{Im} a^{\prime} \geqslant 0$，否则可作变换 $\theta \rightarrow-\theta$
这样对任意的 $\theta\left(0<\theta<\dfrac{\pi}{2}\right)$，有
$\dfrac{1}{r^{2}} \geqslant r^{-2} c^{\prime} \sin 2 \theta+r^{-1} b^{\prime} \sin \theta \geqslant 2 r^{-\frac{3}{2}} \sqrt{b^{\prime} c^{\prime} \sin 2 \theta \cdot \sin \theta}\Rightarrow|b c|=b^{\prime} c^{\prime} \leqslant \dfrac{1}{4 r \sin 2 \theta \cdot \sin \theta}$
（对任意 $r \leqslant 1, \theta \in\left(0, \dfrac{\pi}{2}\right)$）
$\Rightarrow|b c| \leqslant \min\limits _{r \leqslant 1, \theta \in\left(0 , \frac{\pi}{2}\right)} \dfrac{1}{4 r \sin 2 \theta \cdot \sin \theta}=\min\limits _{\theta \in(0 ,\frac{\pi}{2})}\dfrac{1}{4 \sin 2 \theta \cdot \sin \theta}=\dfrac{1}{4 \max \limits_{\theta \in\left(0,\frac{\pi}{2}\right)} \sin 2 \theta \cdot \sin \theta}=\dfrac{3 \sqrt{3}}{16}\cdot|b c|=\dfrac{3 \sqrt{3}}{16}$
的例子：$f(z)=\dfrac{\sqrt{2}}{8} z^{2}-\dfrac{\sqrt{6}}{4} z-\dfrac{3 \sqrt{2}}{8}$ 对于 $z=r \mathrm{e}^{\mathrm{i} \theta}(r \leqslant 1)$，有
$\left|f\left(r \mathrm{e}^{\mathrm{i} \theta}\right)\right|^{2}=\dfrac{1}{32}\left[\left(r^{2} \cos 2 \theta-2 \sqrt{3} r \cos \theta-3\right)^{2}+\right.\left(r^{2} \sin 2 \theta-2 \sqrt{3} r \sin \theta\right)^{2} ]\\=\dfrac{1}{32}\left[2 x^{4}+12 r^{2}+18-\left(2 \sqrt{3} r \cos \theta+r^{2}-3\right)^{2}\right]\leqslant \dfrac{1}{32}\left(2 r^{4}+12 r^{2}+18\right) \leqslant 1$
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