Let's carefully verify Problem 1: $i^9 + 4i^3 - (4 - 2i)$.

Step-by-step Solution:

1. Simplify $i^9$: Recall the cyclic nature of powers of $i$: $i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1 \quad \text{(repeats every 4 powers)}.$ Since $9 \div 4$ leaves a remainder of $1$, we have: $i^9 = i^1 = i.$

2. Simplify $4i^3$: From the cycle above, $i^3 = -i$. Thus: $4i^3 = 4(-i) = -4i.$

3. Expand $-(4 - 2i)$: Distribute the negative sign: $-(4 - 2i) = -4 + 2i.$

4. Combine all terms: Substitute the simplified components: $i^9 + 4i^3 - (4 - 2i) = i - 4i - 4 + 2i.$ Combine like terms:

   • For imaginary parts: $i - 4i + 2i = -i$.
   • For real parts: $-4$.

   Final result: $-4 - i.$

Is the original answer correct?

The student's final answer is written as $-6 - 3i^2$, but this is incorrect. The correct answer is: $\boxed{-4 - i}.$

Would you like a detailed explanation of any step or assistance with the other problems?

Related Questions:

1. How does the cycle of powers of $i$ (i.e., $i, -1, -i, 1$) work?
2. Can you simplify $i^{20} + i^{25}$?
3. What are common mistakes when distributing negative signs in complex numbers?
4. How can we graphically represent $-4 - i$ on the complex plane?
5. What happens if we multiply $-4 - i$ by its complex conjugate?

Tip: Memorizing the cyclic powers of $i$ is key to solving problems involving higher powers of $i$.