To solve \( i^{27} \), we can use the properties of the imaginary unit \( i \).

First, recall the powers of \( i \):
\[
\begin{aligned}
i^1 & = i \\
i^2 & = -1 \\
i^3 & = -i \\
i^4 & = 1 \\
\end{aligned}
\]
These powers repeat every four terms: \( i, -1, -i, 1 \).

To find \( i^{27} \), we can determine the remainder when 27 is divided by 4:

\[
27 \div 4 = 6 \text{ remainder } 3
\]

So, \( i^{27} \) is equivalent to \( i^3 \).

From the list above:

\[
i^3 = -i
\]

Therefore, \( i^{27} = \boxed{-i} \).

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:

1. What is \( i^{50} \)?
2. How does the pattern of powers of \( i \) help in complex number calculations?
3. What is the significance of the imaginary unit \( i \) in complex numbers?
4. How do you solve equations involving \( i \)?
5. What is the geometric interpretation of \( i \) on the complex plane?

**Tip:** When dealing with powers of \( i \), find the remainder when the exponent is divided by 4 to simplify the expression.

Learn how to solve the expression i^27 using the properties of the imaginary unit i. Understand the pattern of powers of i and how to simplify complex number calculations. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation