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Ch. 1 - Equations and Inequalities
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. 1 - Equations and InequalitiesProblem 93
Chapter 2, Problem 93

Simplify each power of i. i23

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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 \(i^4 = 1\).
To simplify \(i^{23}\), find the remainder when 23 is divided by 4, since the powers repeat every 4.
Calculate \(23 \div 4\) which gives a quotient of 5 and a remainder of 3, so \(i^{23} = i^3\).
Use the cycle to identify \(i^3 = -i\), so \(i^{23}\) simplifies to \(-i\).

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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

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 this context. For example, to simplify i^23, compute 23 mod 4 = 3, so i^23 = i^3 = -i. This technique streamlines calculations involving cyclic patterns.
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Arithmetic Sequences - General Formula
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