1. **State the problem:** Calculate $(-\sqrt{3} + i)^7$. 2. **Convert to polar form:** Let $z = -\sqrt{3} + i$. We find the modulus $r$ and argument $\theta$. $$r = |z| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$$ $$\theta = \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right)$$ Since the point is in the second quadrant (negative real, positive imaginary), $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ 3. **Apply De Moivre's theorem:** $$z^7 = r^7 \left(\cos(7\theta) + i \sin(7\theta)\right) = 2^7 \left(\cos\left(7 \times \frac{5\pi}{6}\right) + i \sin\left(7 \times \frac{5\pi}{6}\right)\right)$$ $$= 128 \left(\cos\left(\frac{35\pi}{6}\right) + i \sin\left(\frac{35\pi}{6}\right)\right)$$ 4. **Simplify the angle:** $$\frac{35\pi}{6} = 2\pi \times 2 + \frac{11\pi}{6}$$ Since cosine and sine are periodic with period $2\pi$, $$\cos\left(\frac{35\pi}{6}\right) = \cos\left(\frac{11\pi}{6}\right), \quad \sin\left(\frac{35\pi}{6}\right) = \sin\left(\frac{11\pi}{6}\right)$$ 5. **Evaluate trigonometric functions:** $$\cos\left(\frac{11\pi}{6}\right) = \cos\left(2\pi - \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$ $$\sin\left(\frac{11\pi}{6}\right) = \sin\left(2\pi - \frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$$ 6. **Write the final answer:** $$z^7 = 128 \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = 64 \sqrt{3} - 64 i$$ **Final result:** $$(-\sqrt{3} + i)^7 = 64 \sqrt{3} - 64 i$$