1. Problem (c): Compute $ (2 + 2i)^4$. Note that $2+2i=2(1+i)$. Compute $ (1+i)^2 = 1+2i+i^2 = 2i$. Then $ (1+i)^4 = ((1+i)^2)^2 = (2i)^2 = -4$. Therefore $ (2+2i)^4 = 2^4\cdot(1+i)^4 = 16\cdot -4 = -64$. Answer: $-64$. 2. Problem (d): Compute $ (1 + i)^6$. Note that $ (1+i)^2 = 1+2i+i^2 = 2i$. Then $ (1+i)^6 = ((1+i)^2)^3 = (2i)^3 = 8i^3 = -8i$. Answer: $-8i$. 3. Problem 9: If a, b, c, d are four consecutive integers, evaluate $ i^a + i^b + i^c + i^d$. Let $a=k$ so the sum is $ i^k + i^{k+1} + i^{k+2} + i^{k+3}$. Factor out $i^k$ to get $ i^k(1 + i + i^2 + i^3)$. Compute $ 1 + i + i^2 + i^3 = 1 + i - 1 - i = 0$. Therefore the sum is $ i^k\cdot 0 = 0$. Answer: $0$, choice (a) zero.