Simplify  
i^45
 
Provide your answer below:

To simplify \( i^{45} \), we can use the cyclical property of powers of \( i \), where:

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

Since the powers of \( i \) repeat every 4 terms, we can reduce the exponent \( 45 \) by dividing it by 4 and considering the remainder.

First, calculate the remainder when 45 is divided by 4:

\[
45 \div 4 = 11 \text{ remainder } 1
\]

This means:

\[
i^{45} = i^1 = i
\]

Thus, \( i^{45} = i \).

Would you like more details or have any other questions? Here are five related questions for further practice:

1. What is \( i^{50} \)?
2. Simplify \( i^{100} \).
3. Calculate \( i^{30} \).
4. What is the pattern of \( i^n \) for \( n = 1, 2, 3, 4 \)?
5. Simplify \( (i^2)^5 \).

**Tip:** When dealing with powers of \( i \), always reduce the exponent modulo 4 to find the equivalent simpler power.

Learn how to simplify powers of the imaginary unit i, such as i^45. This problem explains the cyclic nature of i and uses modular arithmetic to simplify complex powers. Suitable for students in Grades 9-12, this explanation includes step-by-step instructions and practice problems.

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