## College Algebra (10th Edition)

Published by Pearson

# Chapter 1 - Section 1.3 - Complex Numbers; Quadratic Equations in the Complex Number System - 1.3 Assess Your Understanding - Page 112: 37

#### Answer

$i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $i^{-15} ,$ use the laws of exponents and the equivalence $i^2=-1.$ $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} i^{-16+1} \\\\= i^{-16}\cdot i^{1} \\\\= i^{-16}\cdot i .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{i^{16}}\cdot i .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{i^{2\cdot8}}\cdot i \\\\= \dfrac{1}{(i^{2})^8}\cdot i .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{(-1)^8}\cdot i \\\\= \dfrac{1}{1}\cdot i \\\\= 1\cdot i \\\\= i .\end{array}

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