b) $z^3 = -i \Leftrightarrow z^3 = e^{\frac{3\pi}{2} i}$ .
Härav

z_k = e^{\frac{i \left(\frac{3\pi}{2} + 2k\pi\right)}{3}}, \quad k = 0,1,2 z_0 = e^{\frac{i\pi}{2}} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} = i z_{1} = e^{\frac{i \left(\frac{3\pi}{2} + 2\pi\right)}{3}} = \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} = (\text{rita trig. cirkeln}) = -\cos \frac{\pi}{6} - i \sin \frac{\pi}{6} = -\frac{\sqrt{3}}{2} - i \frac{1}{2} z_{2} = e^{\frac{i \left(\frac{3\pi}{2} + 4\pi\right)}{3}} = \cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6} = \cos \frac{\pi}{6} - i \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} - i \frac{1}{2}

Svar b) $z_0 = i$ , $z_1 = -\frac{\sqrt{3}}{2} - i \frac{1}{2}$ , $z_2 = \frac{\sqrt{3}}{2} - i \frac{1}{2}$