第11章 11.9

(1)

$z_1 = 4\left(\cos\frac{2}{3}\pi + i\sin\frac{2}{3}\pi\right) = 4\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -2 + i2\sqrt{3}$

$z_2 = 2\left(\cos\frac{1}{6}\pi + i\sin\frac{1}{6}\pi\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i$

$z_1z_2 = (-2 + i2\sqrt{3})(\sqrt{3} + i) = -2\sqrt{3} + i4 - 2\sqrt{3} = -4\sqrt{3} + i4 = 8\left(\cos\frac{5}{6}\pi + i\sin\frac{5}{6}\pi\right)$

$= 8\left(\cos\frac{5}{6}\pi + i\sin\frac{5}{6}\pi\right)$

$\frac{z_1}{z_2} = \frac{-2 + i2\sqrt{3}}{\sqrt{3} + i} = \frac{(-2 + i2\sqrt{3})(\sqrt{3} - i)}{3 + 1} = \frac{-2\sqrt{3} + i8 + 2\sqrt{3}}{4} = 2i = 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$

(2)

$z_1 = 2\left(\cos\frac{5}{4}\pi + i\sin\frac{5}{4}\pi\right) = 2\left(-\frac{\sqrt{2}}{2} + i\left(-\frac{\sqrt{2}}{2}\right)\right) = -\sqrt{2} - i\sqrt{2}$

$z_2 = \cos\frac{5}{12}\pi + i\sin\frac{5}{12}\pi = \cos\left(\frac{2}{12} + \frac{3}{12}\right)\pi + i\sin\left(\frac{2}{12} + \frac{3}{12}\right)\pi$

$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + i\left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}\right) = \frac{1}{4}\{(\sqrt{6} - \sqrt{2}) + i(\sqrt{6} + \sqrt{2})\}$

$z_1z_2 = (-\sqrt{2} - i\sqrt{2}) \cdot \frac{1}{4}\{(\sqrt{6} - \sqrt{2}) + i(\sqrt{6} + \sqrt{2})\} = -\frac{1}{4}(4 + i4\sqrt{3}) = 1 - i\sqrt{3}$

$= 2\left(\cos\frac{5}{3}\pi + i\sin\frac{5}{3}\pi\right)$

$\frac{z_1}{z_2} = \frac{-\sqrt{2} - i\sqrt{2}}{\frac{1}{4}\{(\sqrt{6} - \sqrt{2}) + i(\sqrt{6} + \sqrt{2})\}} = 4\frac{(-\sqrt{2} - i\sqrt{2})\{(\sqrt{6} - \sqrt{2}) - i(\sqrt{6} + \sqrt{2})\}}{(\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} + \sqrt{2})^2}$

$= 4 \cdot \frac{-4\sqrt{3} + 4i}{16} = -\sqrt{3} + i = 2\left(\cos\frac{5}{6}\pi + i\sin\frac{5}{6}\pi\right)$