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Use the defin a) 7+√√=9 2+3i a = 2 b=3 b) √-16 पा I ▬▬▬▬▬▬▬▬ =4 b =

Use the defin a) 7+√√=9 2+3i a = 2 b=3 b) √-16 पा I ▬▬▬▬▬▬▬▬ =4 b =

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Certainly! Below is the transcription and explanation for educational purposes. --- 1. \(x^2 = -1\) has no **real** solutions. But \(x = \pm \sqrt{-1}\) is a solution since \((\sqrt{-1})^2 = -1\) and \((-\sqrt{-1})^2 = -1\). This observation led to the introduction of **complex numbers** and the **imaginary unit** \(i\). \(i\) is defined by \(i^2 = -1\) or \(i = \sqrt{-1}\); complex number \(a + bi\) where \(a\) is real and \(b\) is imaginary. Use the definition and write each number in complex form, \(a + bi\), identifying \(a\) and \(b\). a) \(7 + \sqrt{-9}\) - Solution: - \(7 + 3i\) - \(a = 7\) - \(b = 3\) b) \(\sqrt{-16}\) - Solution: - \(4i\) - \(a = 0\) - \(b = 4\) --- In the problems presented, \(\sqrt{-9}\) and \(\sqrt{-16}\) are examples of numbers expressed using the imaginary unit \(i\). - For part (a): \(7 + \sqrt{-9}\) simplifies to \(7 + 3i\) because \(\sqrt{-9} = 3i\). - For part (b): \(\sqrt{-16}\) simplifies to \(4i\) because \(\sqrt{-16} = 4i\). In both cases, you identify each complex number in the form \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part.
Transcribed Image Text:Certainly! Below is the transcription and explanation for educational purposes. --- 1. \(x^2 = -1\) has no **real** solutions. But \(x = \pm \sqrt{-1}\) is a solution since \((\sqrt{-1})^2 = -1\) and \((-\sqrt{-1})^2 = -1\). This observation led to the introduction of **complex numbers** and the **imaginary unit** \(i\). \(i\) is defined by \(i^2 = -1\) or \(i = \sqrt{-1}\); complex number \(a + bi\) where \(a\) is real and \(b\) is imaginary. Use the definition and write each number in complex form, \(a + bi\), identifying \(a\) and \(b\). a) \(7 + \sqrt{-9}\) - Solution: - \(7 + 3i\) - \(a = 7\) - \(b = 3\) b) \(\sqrt{-16}\) - Solution: - \(4i\) - \(a = 0\) - \(b = 4\) --- In the problems presented, \(\sqrt{-9}\) and \(\sqrt{-16}\) are examples of numbers expressed using the imaginary unit \(i\). - For part (a): \(7 + \sqrt{-9}\) simplifies to \(7 + 3i\) because \(\sqrt{-9} = 3i\). - For part (b): \(\sqrt{-16}\) simplifies to \(4i\) because \(\sqrt{-16} = 4i\). In both cases, you identify each complex number in the form \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part.
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