To solve \((0 + i)^{923}\), we need to understand the properties of complex numbers.

The number \(0 + i\) is a purely imaginary number, and we can represent it as \(i\).

### Step-by-Step Explanation:

1. **Simplify the Expression:**
   \[
   (0 + i)^{923} = i^{923}
   \]

2. **Understanding Powers of \(i\):**
   The powers of \(i\) follow a repeating cycle:
   - \(i^1 = i\)
   - \(i^2 = -1\)
   - \(i^3 = -i\)
   - \(i^4 = 1\)

   This cycle repeats every four terms.

3. **Finding the Equivalent Power Modulo 4:**
   Since the powers of \(i\) repeat every 4, we can reduce the exponent by dividing by 4 and finding the remainder. This helps determine the equivalent power:
   \[
   923 \div 4 = 230 \text{ remainder } 3
   \]
   So, \(i^{923} = i^3\).

4. **Using the Cycle:**
   From the cycle above, we know:
   \[
   i^3 = -i
   \]

### Final Answer:
\[
i^{923} = -i
\]

So, \((0 + i)^{923} = -i\).

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Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. What are the powers of other complex numbers like \(1 + i\)?
2. How would you solve \((0 + i)^n\) for any \(n\)?
3. What is the general rule for powers of \(i\)?
4. How do we simplify higher powers of complex numbers like \(2 + 3i\)?
5. What happens when raising real numbers to powers similar to \(i\)?

**Tip:** Powers of \(i\) repeat every 4 terms, which makes simplifying them easy by using modulo 4.

Understand how to solve the complex number expression (0 + i)^923. This problem uses the properties of powers of i and modulo arithmetic to simplify the exponent. Suitable for 11th grade students, with step-by-step guidance on using the repeating cycle of i^n.

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