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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 67

In Exercises 65–70, perform the indicated operation(s) and write the result in standard form. (2 + i)2 - (3 - i)2

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Step 1: Recall the formula for squaring a binomial: \((a + b)^2 = a^2 + 2ab + b^2\). Apply this formula to \((2 + i)^2\), where \(a = 2\) and \(b = i\). Expand it as \(2^2 + 2(2)(i) + i^2\).
Step 2: Simplify \((2 + i)^2\) by calculating each term: \(2^2 = 4\), \(2(2)(i) = 4i\), and \(i^2 = -1\) (since \(i^2 = -1\) by definition of the imaginary unit). Combine these to get \(4 + 4i - 1\).
Step 3: Simplify \(4 + 4i - 1\) to \(3 + 4i\). This is the result of \((2 + i)^2\).
Step 4: Similarly, apply the formula for squaring a binomial to \((3 - i)^2\), where \(a = 3\) and \(b = -i\). Expand it as \(3^2 + 2(3)(-i) + (-i)^2\).
Step 5: Simplify \((3 - i)^2\) by calculating each term: \(3^2 = 9\), \(2(3)(-i) = -6i\), and \((-i)^2 = -1\). Combine these to get \(9 - 6i - 1\), which simplifies to \(8 - 6i\). Subtract \((3 - i)^2\) from \((2 + i)^2\) to get \((3 + 4i) - (8 - 6i)\), and simplify the real and imaginary parts separately.

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Complex numbers are numbers that have a real part and an imaginary part, expressed in the form a + bi, where a is the real part, b is the imaginary part, and i is the imaginary unit defined as the square root of -1. Understanding how to manipulate complex numbers is essential for performing operations such as addition, subtraction, multiplication, and division.
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