1. **State the problem:** Find the value of $ (1 - i)^4 $ using De Moivre's theorem. 2. **Recall De Moivre's theorem:** For a complex number in polar form $ r(\cos \theta + i \sin \theta) $, its $n$th power is given by: $$ (r(\cos \theta + i \sin \theta))^n = r^n (\cos n\theta + i \sin n\theta) $$ 3. **Convert $1 - i$ to polar form:** - Calculate the modulus: $$ r = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} $$ - Calculate the argument $\theta$: $$ \theta = \arctan\left(\frac{-1}{1}\right) = \arctan(-1) = -\frac{\pi}{4} $$ 4. **Apply De Moivre's theorem:** $$ (1 - i)^4 = (\sqrt{2})^4 \left( \cos 4\left(-\frac{\pi}{4}\right) + i \sin 4\left(-\frac{\pi}{4}\right) \right) $$ 5. **Simplify powers and angles:** $$ (\sqrt{2})^4 = (\sqrt{2}^2)^2 = (2)^2 = 4 $$ $$ 4 \times \left(-\frac{\pi}{4}\right) = -\pi $$ 6. **Evaluate trigonometric functions:** $$ \cos(-\pi) = \cos \pi = -1 $$ $$ \sin(-\pi) = -\sin \pi = 0 $$ 7. **Write the final result:** $$ (1 - i)^4 = 4(-1 + 0i) = -4 $$ **Answer:** $ (1 - i)^4 = -4 $