1. **State the problem:** Find the cube roots of the complex number $i$. 2. **Formula and explanation:** To find the cube roots of a complex number, express it in polar form $z = r(\cos \theta + i \sin \theta)$, then use De Moivre's theorem: $$\sqrt[3]{z} = r^{1/3} \left( \cos \frac{\theta + 2k\pi}{3} + i \sin \frac{\theta + 2k\pi}{3} \right)$$ where $k = 0, 1, 2$ for the three cube roots. 3. **Convert $i$ to polar form:** - The complex number $i$ is $0 + 1i$. - Its magnitude is $r = |i| = 1$. - Its argument (angle) is $\theta = \frac{\pi}{2}$ because it lies on the positive imaginary axis. 4. **Calculate cube roots:** - Magnitude of cube roots: $r^{1/3} = 1^{1/3} = 1$. - Angles for cube roots: - For $k=0$: $\frac{\pi/2 + 2\cdot0\cdot\pi}{3} = \frac{\pi}{6}$ - For $k=1$: $\frac{\pi/2 + 2\cdot1\cdot\pi}{3} = \frac{\pi/2 + 2\pi}{3} = \frac{5\pi}{6}$ - For $k=2$: $\frac{\pi/2 + 2\cdot2\cdot\pi}{3} = \frac{\pi/2 + 4\pi}{3} = \frac{9\pi}{6} = \frac{3\pi}{2}$ 5. **Write cube roots in rectangular form:** - For $k=0$: $$\cos \frac{\pi}{6} + i \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$$ - For $k=1$: $$\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i$$ - For $k=2$: $$\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} = 0 - 1i = -i$$ 6. **Final answer:** The three cube roots of $i$ are $$\frac{\sqrt{3}}{2} + \frac{1}{2}i, \quad -\frac{\sqrt{3}}{2} + \frac{1}{2}i, \quad -i.$$