Simplify each power of i. i²⁷

Verified step by step guidance

1. Recall that the imaginary unit \(i\) is defined such that \(i^2 = -1\).
2. Recognize that powers of \(i\) repeat in a cycle of 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\).
3. To simplify \(i^{27}\), find the remainder when 27 is divided by 4, since the powers repeat every 4.
4. Calculate \(27 \div 4\) which gives a quotient of 6 and a remainder of 3, so \(27 \equiv 3 \pmod{4}\).
5. Therefore, \(i^{27} = i^3\), and from the cycle, \(i^3 = -i\).

Key Concepts

Imaginary Unit (i)

The imaginary unit i is defined as the square root of -1, satisfying i² = -1. It is the foundation of complex numbers and allows for the extension of real numbers to include solutions to equations like x² + 1 = 0.

Powers of i and Their Cyclic Pattern

Powers of i repeat in a cycle of four: i¹ = i, i² = -1, i³ = -i, and i⁴ = 1. This pattern repeats for higher powers, so simplifying i raised to any integer power involves finding the remainder when the exponent is divided by 4.

Modular Arithmetic for Exponent Simplification

Modular arithmetic helps simplify powers by reducing the exponent modulo 4 in this context. For example, to simplify i^27, compute 27 mod 4 = 3, so i^27 = i³ = -i, making complex exponentiation manageable.