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Ch. 1 - Equations and Inequalities
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 2, Problem 92

Simplify each power of i. i26

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1
Recall that the imaginary unit \(i\) is defined such that \(i^2 = -1\).
Recognize that powers of \(i\) repeat in a cycle of 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\), then the pattern repeats.
To simplify \(i^{26}\), find the remainder when 26 is divided by 4, since the powers cycle every 4 steps.
Calculate \(26 \div 4\) which gives a quotient of 6 and a remainder of 2, so \(i^{26} = i^{4 \times 6 + 2} = (i^4)^6 \times i^2\).
Since \(i^4 = 1\), simplify to \(1^6 \times i^2 = i^2\), and recall that \(i^2 = -1\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Imaginary Unit i

The imaginary unit i is defined as the square root of -1, satisfying i² = -1. It is the fundamental unit used to extend the real number system to complex numbers, allowing for the representation and manipulation of numbers involving the square roots of negative values.
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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.
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Modular Arithmetic for Exponent Simplification

Modular arithmetic helps simplify powers by reducing the exponent modulo 4 in the case of i. For example, to simplify i^26, compute 26 mod 4 = 2, so i^26 = i² = -1. This technique streamlines calculations involving cyclic patterns.
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