Simplify each power of i. i25

Ch. 1 - Equations and Inequalities

Lial13th EditionCollege AlgebraISBN: 9780136881063당신이 사용하는 게 아니라요?교과서 변경

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Lial 13th Edition

Ch. 1 - Equations and Inequalities

문제 89a

2장, 문제 89a

Simplify each power of i. i²⁵

검증된 단계별 안내

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 \(i^{25}\), find the remainder when 25 is divided by 4, since the powers repeat every 4.

Calculate \(25 \div 4\) which gives a quotient of 6 and a remainder of 1, so \(i^{25} = i^{4 \cdot 6 + 1} = (i^4)^6 \cdot i^1\).

Use the fact that \((i^4)^6 = 1^6 = 1\), so \(i^{25} = 1 \cdot i = i\).

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주요 개념

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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 of numbers involving the square roots of negative values.

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^25, compute 25 mod 4 = 1, so i^25 = i¹ = i. This technique streamlines calculations involving cyclic patterns.