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.