1. The problem is to find the sum of the series $$1 + i + i^2 + i^3 + i^4 + \cdots + i^{2021}$$ where $i$ is the imaginary unit with the property $i^2 = -1$. 2. Recall the powers of $i$ cycle every 4 terms: $$i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \ldots$$ 3. Since the powers repeat every 4, the sum can be grouped into blocks of 4 terms: $$S = (1 + i + i^2 + i^3) + (i^4 + i^5 + i^6 + i^7) + \cdots + i^{2021}$$ 4. Each block of 4 terms sums to zero: $$1 + i + (-1) + (-i) = 0$$ 5. Find how many complete blocks of 4 terms are in 2022 terms (from $i^0$ to $i^{2021}$): $$\text{Number of blocks} = \left\lfloor \frac{2022}{4} \right\rfloor = 505$$ 6. The sum of these 505 blocks is: $$505 \times 0 = 0$$ 7. Now, find the remainder terms after these blocks: $$2022 - 505 \times 4 = 2022 - 2020 = 2$$ 8. The remainder terms are $i^{2020}$ and $i^{2021}$. 9. Calculate these remainder terms using the cycle: $$i^{2020} = i^{4 \times 505} = (i^4)^{505} = 1^{505} = 1$$ $$i^{2021} = i^{2020} \times i = 1 \times i = i$$ 10. Sum of remainder terms: $$1 + i$$ 11. Therefore, the total sum is: $$S = 0 + (1 + i) = 1 + i$$ Final answer: $$\boxed{1 + i}$$