填充第7題
設\(z\)為複數，若\(|\;z|\;=2\)，則\(|\;z^2-2z+8|\;\)的最小值為
。
(類似問題，
https://math.pro/db/viewthread.php?tid=1143&page=1#pid3600)
[解答]
令\(z=2\left( \cos \theta +i\sin \theta \right)\)
\(\begin{align}
& \left| {{z}^{2}}-2z+8 \right| \\
& =\left| z \right|\left| z-2+\frac{8}{z} \right| \\
& =2\left| 2\cos \theta +2i\sin \theta -2+8\left( \frac{\cos \left( -\theta \right)+i\sin \left( -\theta \right)}{2} \right) \right| \\
& =2\left| 6\cos \theta -2-2i\sin \theta \right| \\
& =2\sqrt{{{\left( 6\cos \theta -2 \right)}^{2}}+{{\left( -2\sin \theta \right)}^{2}}} \\
& =2\sqrt{32{{\cos }^{2}}\theta -24\cos \theta +8} \\
& =2\sqrt{32{{\left( \cos \theta -\frac{3}{8} \right)}^{2}}+\frac{7}{2}} \\
& \ge \sqrt{14} \\
\end{align}\)
