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if 𝒏 is an integer, write each of the following in the simplest form

The expression given is \( i^{4n + 25} \), where \( n \) is an integer.

To simplify this expression, we need to make use of the cyclic nature of powers of \( i \), the imaginary unit. Recall that:

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

This cycle repeats every 4 powers. So to simplify \( i^{4n + 25} \), we need to reduce the exponent \( 4n + 25 \) modulo 4.

### Step 1: Simplify the exponent \( 4n + 25 \mod 4 \)
\[
4n + 25 \equiv 25 \mod 4
\]

Now, calculate \( 25 \mod 4 \):
\[
25 \div 4 = 6 \text{ remainder } 1
\]
Thus,
\[
25 \mod 4 = 1
\]

### Step 2: Apply the cyclic property of powers of \( i \)
Since \( 4n + 25 \equiv 1 \mod 4 \), we have:
\[
i^{4n + 25} = i^1 = i
\]

### Final Answer:
\[
i^{4n + 25} = i
\]

Would you like a detailed explanation on the powers of \( i \) or further clarifications?

Here are some related questions:
1. What are the general rules for simplifying powers of complex numbers?
2. How do you simplify expressions like \( i^{100} \)?
3. Why does the cycle for powers of \( i \) repeat every four steps?
4. Can powers of \( i \) be applied to solve real-world problems involving imaginary numbers?
5. How does \( i^2 = -1 \) impact equations in algebra and calculus?

**Tip:** Always reduce powers of imaginary numbers using modular arithmetic with respect to 4 to simplify quickly!

If n is an integer, write i^(4n+25) in its simplest form.

Learn how to simplify the expression i^(4n+25) using the cyclic nature of powers of i. This problem covers key concepts in complex numbers and modular arithmetic, suitable for students in Grades 9-12.

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