第11章 11.10

(1)

$z = 1 + 3i$ $\theta = \frac{\pi}{3}$

$\cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$

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

(2)

$z = 4 - i$ $\theta = \frac{3}{4}\pi$

$\cos\frac{3}{4}\pi + i\sin\frac{3}{4}\pi = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$

$(4-i)\left(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = -2\sqrt{2} + \frac{\sqrt{2}}{2} + 2\sqrt{2}i + \frac{\sqrt{2}}{2}i = -\frac{3\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}i$

(3)

$z = 2 + 8i$ $\theta = \frac{5}{6}\pi$

$\cos\frac{5}{6}\pi + i\sin\frac{5}{6}\pi = -\frac{\sqrt{3}}{2} + i\frac{1}{2}$

$(2+8i)\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} - 4i - 4\sqrt{3} = -4 - \sqrt{3} + (1-4\sqrt{3})i$

(4)

$z = -3 + 2i$ $\theta = -\frac{1}{6}\pi$

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

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