1. **Problem statement:** Find the two values of $$\sqrt{2(1 - \sqrt{3}i)}$$ using De Moivre's theorem, expressing each solution in the form $$a + bi$$ where $$a,b \in \mathbb{R}$$ and $$i^2 = -1$$. 2. **Rewrite the complex number:** Let $$z = 2(1 - \sqrt{3}i) = 2 - 2\sqrt{3}i$$. 3. **Convert to polar form:** - Calculate modulus $$r = |z| = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 4 \times 3} = \sqrt{4 + 12} = \sqrt{16} = 4$$. - Calculate argument $$\theta = \tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$$ (since real part positive and imaginary part negative, angle is in 4th quadrant). 4. **Apply De Moivre's theorem for square roots:** The two square roots are given by: $$w_k = r^{1/2} \left( \cos\left(\frac{\theta + 2k\pi}{2}\right) + i \sin\left(\frac{\theta + 2k\pi}{2}\right) \right), \quad k=0,1$$ 5. **Calculate $$r^{1/2}$$:** $$r^{1/2} = \sqrt{4} = 2$$. 6. **Calculate roots:** - For $$k=0$$: $$\frac{\theta}{2} = \frac{-\pi/3}{2} = -\frac{\pi}{6}$$ $$w_0 = 2 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) = 2 \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = \sqrt{3} - i$$ - For $$k=1$$: $$\frac{\theta + 2\pi}{2} = \frac{-\pi/3 + 2\pi}{2} = \frac{5\pi/3}{2} = \frac{5\pi}{6}$$ $$w_1 = 2 \left( \cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right) \right) = 2 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\sqrt{3} + i$$ 7. **Final answers:** $$\boxed{\sqrt{3} - i \quad \text{and} \quad -\sqrt{3} + i}$$ These are the two square roots of $$2(1 - \sqrt{3}i)$$ in the form $$a + bi$$.