Problem 58
Question
Multiply as indicated. Write each product in standard form. $$(1+3 i)(2-5 i)$$
Step-by-Step Solution
Verified Answer
The product in standard form is \(17 + i\).
1Step 1: 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.
2Step 2: Calculate Each Term
Calculate each multiplication:- \(1 \times 2 = 2\),- \(1 \times (-5i) = -5i\),- \(3i \times 2 = 6i\),- \(3i \times (-5i) = -15i^2\).
3Step 3: Simplify Using \(i^2 = -1\)
Remember that \(i^2 = -1\). Substituting, we have:\(-15i^2 = -15(-1) = 15.\)
4Step 4: Combine Like Terms
Combine the real parts and the imaginary parts from the equation:\(2 + 15 + (-5i + 6i) = 17 + i.\)
5Step 5: 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:\(17 + i\).
Key Concepts
Distributive PropertyFOIL MethodImaginary Unit
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:
By distributing each term across the other parentheses, we take:
- \(1 \times 2\)
- \(1 \times (-5i)\)
- \(3i \times 2\)
- \(3i \times (-5i)\)
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 \(3i \times (-5i)\)
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.
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.
Other exercises in this chapter
Problem 58
RELATING CONCEPTS For individual or group investigation (Exercises \(55-60\) ) The close relationships among \(x\) -intercepts of a graph of a function, real ze
View solution Problem 58
Find the equation of the quadratic function satisfying the given conditions. (Hint: Determine values of \(a\), \(h,\) and \(k\) that satisfy \(P(x)=a(x-h)^{2}+k
View solution Problem 58
Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{3}=-8$$
View solution Problem 58
The concepts of stretching, shrinking, translating, and reflecting graphs presented in Sections 2.2 and 2.3 can be applied to polynomial functions of the form \
View solution