1. **State the problem:** Find the roots of the complex equation $$z^3 + 27 \operatorname{cis}\frac{3\pi}{4} = 0$$ where $$\operatorname{cis}\theta = \cos\theta + i\sin\theta$$. 2. **Rewrite the equation:** Move the constant term to the right side: $$z^3 = -27 \operatorname{cis}\frac{3\pi}{4}$$ 3. **Express the right side in polar form:** Note that $$-27 = 27 \operatorname{cis}\pi$$, so $$-27 \operatorname{cis}\frac{3\pi}{4} = 27 \operatorname{cis}\left(\pi + \frac{3\pi}{4}\right) = 27 \operatorname{cis}\frac{7\pi}{4}$$ 4. **Use De Moivre's theorem:** The cube roots of a complex number $$r \operatorname{cis}\theta$$ are given by $$z_k = \sqrt[3]{r} \operatorname{cis}\left(\frac{\theta + 2k\pi}{3}\right), \quad k=0,1,2$$ 5. **Calculate the magnitude and arguments:** - Magnitude: $$\sqrt[3]{27} = 3$$ - Arguments: $$\frac{7\pi/4 + 2k\pi}{3} = \frac{7\pi}{12} + \frac{2k\pi}{3}$$ 6. **Find the three roots:** - For $$k=0$$: $$z_0 = 3 \operatorname{cis} \frac{7\pi}{12}$$ - For $$k=1$$: $$z_1 = 3 \operatorname{cis} \left(\frac{7\pi}{12} + \frac{2\pi}{3}\right) = 3 \operatorname{cis} \frac{15\pi}{12} = 3 \operatorname{cis} \frac{5\pi}{4}$$ - For $$k=2$$: $$z_2 = 3 \operatorname{cis} \left(\frac{7\pi}{12} + \frac{4\pi}{3}\right) = 3 \operatorname{cis} \frac{23\pi}{12}$$ 7. **Final answer:** The roots are $$z_0 = 3 \left(\cos \frac{7\pi}{12} + i \sin \frac{7\pi}{12}\right),$$ $$z_1 = 3 \left(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}\right),$$ $$z_2 = 3 \left(\cos \frac{23\pi}{12} + i \sin \frac{23\pi}{12}\right).$$