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.