第11章 11.31

(1)

$(1+2i)^3 = (1+4i-4)(1+2i) = (-3+4i)(1+2i) = -3-2i-8 = -11-2i$

(2)

$\sqrt{i} = z$ とおくと両辺を 2 乗して

$z^2 = i$

$z = r(\cos\theta + i\sin\theta)$ におくと

$r^2(\cos\theta + i\sin\theta)^2 = i \quad r^2(\cos 2\theta + i\sin 2\theta) = i$

これから

$r = 1 \quad 2\theta = \frac{\pi}{2} + 2k\pi \quad \theta = \frac{\pi}{4} + k\pi$

$k=0$ のとき $\theta = \frac{\pi}{4}$、$k=1$ のとき $\theta = \frac{5}{4}\pi$ これから

$z_1 = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$

$z_2 = \cos\frac{5}{4}\pi + i\sin\frac{5}{4}\pi = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i$