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The expression shown is \( i^{23} \), where \( i \) is the imaginary unit defined as \( i = \sqrt{-1} \). 

To simplify this expression, we need to use the property that the powers of \( i \) repeat in cycles of 4:

\[
i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1
\]

After \( i^4 \), the powers repeat this cycle: 

\[
i^5 = i, \quad i^6 = -1, \quad i^7 = -i, \quad i^8 = 1, \ldots
\]

So, to simplify \( i^{23} \), we need to find the remainder when 23 is divided by 4 (because the powers of \( i \) repeat every 4 terms).

\[
23 \div 4 = 5 \text{ remainder } 3
\]

Therefore, \( i^{23} = i^3 \).

From the cycle of powers:

\[
i^3 = -i
\]

So, \( i^{23} = -i \).

### Final answer:
\[
i^{23} = -i
\]

Let me know if you'd like further details or clarifications! 

Here are 5 questions to extend your understanding:
1. What is \( i^{30} \)?
2. How do powers of imaginary numbers affect their values?
3. Can you simplify \( i^{50} \)?
4. Why does the power of \( i \) repeat every 4 terms?
5. What happens when you raise \( i \) to a negative exponent?

**Tip:** When dealing with powers of \( i \), always reduce the exponent modulo 4 to simplify the expression.

Learn how to simplify the expression i^23. This problem uses key concepts of complex numbers and the cyclical nature of powers of the imaginary unit i, which repeat every 4 terms. The solution is broken down for students in Grades 10-12.

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