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Chapter 3: Problem 58

Multiply as indicated. Write each product in standard form. $$(1+3 i)(2-5 i)$$

Short Answer

Expert verified
The product in standard form is \(17 + i\).

Step by step solution

01

Use the Distributive Property

Apply the distributive property, also known as the FOIL method, to expand the expression \((1+3i)(2-5i) = 1 imes 2 + 1 imes (-5i) + 3i imes 2 + 3i imes (-5i).\)Calculate each term separately.
02

Calculate Each Term

Calculate each multiplication:- \(1 \times 2 = 2\),- \(1 \times (-5i) = -5i\),- \(3i \times 2 = 6i\),- \(3i \times (-5i) = -15i^2\).
03

Simplify Using \(i^2 = -1\)

Remember that \(i^2 = -1\). Substituting, we have:\(-15i^2 = -15(-1) = 15.\)
04

Combine Like Terms

Combine the real parts and the imaginary parts from the equation:\(2 + 15 + (-5i + 6i) = 17 + i.\)
05

Write in Standard Form

Ensure the expression is written in standard form \(a + bi\), where \(a\) and \(b\) are real numbers. Here, the standard form is:\(17 + i\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distributive Property
When working with complex numbers, the Distributive Property plays a crucial role. This property allows us to multiply each term in one expression by each term in another. It is represented by the formula: \((a+b)(c+d) = ac+ad+bc+bd\). For our exercise, this means we expand \((1+3i)(2-5i)\).

By distributing each term across the other parentheses, we take:
  • \(1 \times 2\)
  • \(1 \times (-5i)\)
  • \(3i \times 2\)
  • \(3i \times (-5i)\)
These four distributions lead us to the final multiplication step, transforming complex multiplication into a manageable array of simpler multiplications.
FOIL Method
The FOIL Method is often an easy way to remember how to multiply two binomials, especially when handling complex numbers like \((1+3i)(2-5i)\). FOIL stands for:
  • First: Multiply the first terms \(1 \times 2\)
  • Outer: Multiply the outer terms \(1 \times (-5i)\)
  • Inner: Multiply the inner terms \(3i \times 2\)
  • Last: Multiply the last terms \(3i \times (-5i)\)
This method helps ensure no combinations are missed. In our problem, using FOIL method, it breaks down into these four parts which ultimately gives us the full expanded form of the product.
Imaginary Unit
The imaginary unit, denoted as \(i\), is a fundamental concept in complex numbers. Its most essential property is that \(i^2 = -1\). This is important when multiplying terms that include \(i\).

In our exercise, when calculating \(3i \times (-5i)\), we find \(-15i^2\). Since \(i^2 = -1\), we substitute to get \(-15 \times -1 = 15\).

This simplification converts what initially looks like a more challenging problem into a combination of real and imaginary components, ultimately giving us the result in standard form \(a + bi\). Understanding \(i^2 = -1\) is key to simplifying and accurately working through these calculations.

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Most popular questions from this chapter

Divide. $$\frac{2 x^{4}-x^{3}+4 x^{2}+8 x+7}{2 x^{2}+3 x+2}$$

Divide. $$\frac{20 x^{4}+6 x^{3}-2 x^{2}+15 x-2}{5 x-1}$$

Sketch a graph of a quadratic function that satisfies each set of given conditions. Use symmetry to label another point on your graph. Maximum value of 1 at \(x=3 ; y\) -intercept is \((0,-4)\)

Use the concepts of this section. Determine whether the description of the polynomial function \(P(x)\) with real coefficients is possible or not possible. (a) \(P(x)\) is of degree 3 and has zeros of \(1,2,\) and \(1+i\). (b) \(P(x)\) is of degree 4 and has four nonreal complex zeros. (c) \(P(x)\) is of degree 5 and \(-6\) is a zero of multiplicity 6. (d) \(P(x)\) has \(1+2 i\) as a zero of multiplicity 2.

Use the concepts of this section. Explain why a polynomial function of degree 4 with real coefficients has either zero, two, or four real zeros (counting multiplicities).
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