To simplify $i^{32}$, we need to use the cyclical property of powers of the imaginary unit $i$. Recall that:

$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$

This pattern repeats every four powers: $i^5 = i$, $i^6 = -1$, and so on.

Thus, to simplify $i^{32}$, we can divide 32 by 4 and examine the remainder:

$32 \div 4 = 8 \text{ remainder } 0$

Since the remainder is 0, we conclude that:

$i^{32} = i^0 = 1$

Thus, $i^{32} = 1$.

Would you like more details or have any questions?

Here are 5 related questions for deeper understanding:

1. What is the value of $i^{33}$?
2. How does the cyclical nature of powers of $i$ help in complex number calculations?
3. What is the geometric interpretation of $i$ on the complex plane?
4. How do powers of $i$ relate to Euler's formula?
5. Can you simplify $i^{40}$ using the same method?

Tip: Always look for patterns in powers of complex numbers to simplify calculations quickly.