1. **Problem statement:** Express $ (1+i)^8 $ in the form $ a+bi $. 2. **Formula and rules:** Use the polar form of complex numbers and De Moivre's theorem: $$ (r(\cos \theta + i \sin \theta))^n = r^n (\cos n\theta + i \sin n\theta) $$ where $ r = |1+i| $ and $ \theta = \arg(1+i) $. 3. **Calculate modulus and argument:** $$ r = \sqrt{1^2 + 1^2} = \sqrt{2} $$ $$ \theta = \tan^{-1}(\frac{1}{1}) = \frac{\pi}{4} $$ 4. **Apply De Moivre's theorem:** $$ (1+i)^8 = (\sqrt{2})^8 \left( \cos 8 \times \frac{\pi}{4} + i \sin 8 \times \frac{\pi}{4} \right) $$ $$ = (\sqrt{2})^8 (\cos 2\pi + i \sin 2\pi) $$ 5. **Simplify powers and trigonometric values:** $$ (\sqrt{2})^8 = (2^{1/2})^8 = 2^{4} = 16 $$ $$ \cos 2\pi = 1, \quad \sin 2\pi = 0 $$ 6. **Final expression:** $$ (1+i)^8 = 16 (1 + 0i) = 16 $$ **Answer:** $16 + 0i$ or simply $16$.