第11章 11.11

(1)

$(\sqrt{3}+i)^6 = \left\{2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\right\}^6 = 64(\cos\pi + i\sin\pi) = -64$

(2)

$(1+i)^{10} = \left\{\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right\}^{10} = 32\left(\cos\frac{5}{2}\pi + i\sin\frac{5}{2}\pi\right) = 32i$

(3)

$(1-\sqrt{3}i)^6 = \left\{2\left(\cos\frac{5}{3}\pi + i\sin\frac{5}{3}\pi\right)\right\}^6 = 64(\cos 10\pi + i\sin 10\pi) = 64$

(4)

$(1-i)^{-10} = \left\{\sqrt{2}\left(\cos\frac{7}{4}\pi + i\sin\frac{7}{4}\pi\right)\right\}^{-10} = \frac{1}{32}\left\{\cos\left(-\frac{35}{2}\pi\right) + i\sin\left(-\frac{35}{2}\pi\right)\right\} = \frac{1}{32}i$

(5)

$(-1+\sqrt{3}i)^{-4} = \left\{2\left(\cos\frac{2}{3}\pi + i\sin\frac{2}{3}\pi\right)\right\}^{-4} = \frac{1}{16}\left\{\cos\left(-\frac{8}{3}\pi\right) + i\sin\left(-\frac{8}{3}\pi\right)\right\} = -\frac{1}{32}-\frac{\sqrt{3}}{32}i$

(6)

$(\sqrt{3}-i)^{-6} = \left\{2\left(\cos\frac{11}{6}\pi + i\sin\frac{11}{6}\pi\right)\right\}^{-6} = \frac{1}{64}\{\cos(-11\pi) + i\sin(-11\pi)\} = -\frac{1}{64}$