1. **State the problem:** Express the complex number $-8 - 8\sqrt{3}i$ in polar form $r(\cos\theta + i\sin\theta)$, then use de Moivre's theorem to find the four roots of the equation $z^4 = -8 - 8\sqrt{3}i$. 2. **Convert to polar form:** - Calculate the modulus $r = |z| = \sqrt{(-8)^2 + (-8\sqrt{3})^2} = \sqrt{64 + 64 \times 3} = \sqrt{64 + 192} = \sqrt{256} = 16$. - Calculate the argument $\theta = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{-8\sqrt{3}}{-8}\right) = \tan^{-1}(\sqrt{3})$. - Since both real and imaginary parts are negative, the complex number lies in the third quadrant, so $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$. 3. **Polar form:** $$-8 - 8\sqrt{3}i = 16 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right)$$ 4. **Use de Moivre's theorem:** - The equation $z^4 = 16 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right)$. - The fourth roots are given by: $$z_k = r^{1/4} \left( \cos \frac{\theta + 2k\pi}{4} + i \sin \frac{\theta + 2k\pi}{4} \right), \quad k=0,1,2,3$$ - Calculate $r^{1/4} = 16^{1/4} = 2$. 5. **Calculate each root:** - For $k=0$: $$z_0 = 2 \left( \cos \frac{4\pi/3 + 0}{4} + i \sin \frac{4\pi/3 + 0}{4} \right) = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = 2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 1 + i\sqrt{3}$$ - For $k=1$: $$z_1 = 2 \left( \cos \frac{4\pi/3 + 2\pi}{4} + i \sin \frac{4\pi/3 + 2\pi}{4} \right) = 2 \left( \cos \frac{10\pi}{12} + i \sin \frac{10\pi}{12} \right) = 2 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right) = 2 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\sqrt{3} + i$$ - For $k=2$: $$z_2 = 2 \left( \cos \frac{4\pi/3 + 4\pi}{4} + i \sin \frac{4\pi/3 + 4\pi}{4} \right) = 2 \left( \cos \frac{16\pi}{12} + i \sin \frac{16\pi}{12} \right) = 2 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = 2 \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = -1 - i\sqrt{3}$$ - For $k=3$: $$z_3 = 2 \left( \cos \frac{4\pi/3 + 6\pi}{4} + i \sin \frac{4\pi/3 + 6\pi}{4} \right) = 2 \left( \cos \frac{22\pi}{12} + i \sin \frac{22\pi}{12} \right) = 2 \left( \cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6} \right) = 2 \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = \sqrt{3} - i$$ 6. **Final answers:** $$z_0 = 1 + i\sqrt{3}, \quad z_1 = -\sqrt{3} + i, \quad z_2 = -1 - i\sqrt{3}, \quad z_3 = \sqrt{3} - i$$