Simplify each power of i. i27

Ch. 1 - Equations and Inequalities

Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook

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Lial 13th Edition

Ch. 1 - Equations and Inequalities

Problem 94

Chapter 2, Problem 94

Simplify each power of i. i²⁷

Verified step by step guidance

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 $i^{27}$, find the remainder when 27 is divided by 4, since the powers repeat every 4.

Calculate $27 \div 4$ which gives a quotient of 6 and a remainder of 3, so $27 \equiv 3 \pmod{4}$.

Therefore, $i^{27} = i^3$, and from the cycle, $i^3 = -i$.

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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 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.

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