1. **State the problem:** Find the cube root of the complex number $z = -4\sqrt{2} + 4\sqrt{2}i$. 2. **Convert to polar form:** A complex number $z = x + yi$ can be written as $z = r(\cos \theta + i \sin \theta)$ where $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}(\frac{y}{x})$. Calculate $r$: $$r = \sqrt{(-4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8$$ Calculate $\theta$: $$\theta = \tan^{-1}\left(\frac{4\sqrt{2}}{-4\sqrt{2}}\right) = \tan^{-1}(-1)$$ Since $x$ is negative and $y$ is positive, $\theta$ is in the second quadrant: $$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$ So, $$z = 8 \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$$ 3. **Use De Moivre's theorem to find cube roots:** The cube roots of $z$ are given by: $$w_k = r^{1/3} \left( \cos \frac{\theta + 2k\pi}{3} + i \sin \frac{\theta + 2k\pi}{3} \right), \quad k=0,1,2$$ Calculate $r^{1/3}$: $$r^{1/3} = 8^{1/3} = 2$$ Calculate the roots: - For $k=0$: $$w_0 = 2 \left( \cos \frac{3\pi/4}{3} + i \sin \frac{3\pi/4}{3} \right) = 2 \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 2 \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 2 \cdot \frac{\sqrt{2}}{2} (1 + i) = \sqrt{2} (1 + i)$$ - For $k=1$: $$w_1 = 2 \left( \cos \frac{3\pi/4 + 2\pi}{3} + i \sin \frac{3\pi/4 + 2\pi}{3} \right) = 2 \left( \cos \frac{11\pi}{12} + i \sin \frac{11\pi}{12} \right)$$ - For $k=2$: $$w_2 = 2 \left( \cos \frac{3\pi/4 + 4\pi}{3} + i \sin \frac{3\pi/4 + 4\pi}{3} \right) = 2 \left( \cos \frac{19\pi}{12} + i \sin \frac{19\pi}{12} \right)$$ 4. **Final answer:** The three cube roots of $z$ are: $$\boxed{\sqrt{2}(1 + i), \quad 2 \left( \cos \frac{11\pi}{12} + i \sin \frac{11\pi}{12} \right), \quad 2 \left( \cos \frac{19\pi}{12} + i \sin \frac{19\pi}{12} \right)}$$