To simplify \(i^{59}\), we can use the properties of powers of the imaginary unit \(i\).

### Key Property of \(i\)

The imaginary unit \(i\) satisfies:

\[
i^2 = -1
\]

From this, we can find a repeating pattern for powers of \(i\):

\[
\begin{align*}
i^1 &= i, \\
i^2 &= -1, \\
i^3 &= -i, \\
i^4 &= 1.
\end{align*}
\]

After \(i^4 = 1\), the powers repeat every 4 steps. This means the powers of \(i\) form a cycle of 4: \(i, -1, -i, 1, i, -1, -i, 1, \ldots\).

### Simplifying \(i^{59}\)

To simplify \(i^{59}\), we find the remainder when 59 is divided by 4, since the powers repeat every 4.

\[
59 \div 4 = 14 \text{ remainder } 3
\]

So, \(59 \equiv 3 \pmod{4}\). This means:

\[
i^{59} = i^3
\]

From the pattern we know:

\[
i^3 = -i
\]

### Final Answer

\[
i^{59} = -i
\]

Would you like to explore more about powers of \(i\) or have any questions?

Here are 5 related questions to deepen your understanding:

1. How would you simplify \(i^{101}\)?
2. Why does the power of \(i\) repeat every 4 terms?
3. How do complex exponents differ from real exponents?
4. Can you visualize the powers of \(i\) on the complex plane?
5. What happens if you raise \(i\) to a negative power?

**Tip:** Remember, finding the remainder when dividing by 4 is the key to simplifying powers of \(i\).

Learn how to simplify i^59 using the properties of the imaginary unit i. Understand the cyclic nature of powers of i and how to find the remainder when dividing the exponent by 4. Suitable for advanced high school students studying complex numbers.

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