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Chapter 8: Problem 56

Find the following products. $$5 i(3+4 i)$$

Short Answer

Expert verified
The product is \(-20 + 15i\).

Step by step solution

01

Distribute the Multiplication

To find the product, we will distribute the multiplication. Multiply each term in the parentheses by the complex number outside:\[ 5i(3 + 4i) = 5i \cdot 3 + 5i \cdot 4i \]
02

Multiply Real and Imaginary Parts

Perform the multiplication on each distributed term individually:\[ 5i \cdot 3 = 15i \]\[ 5i \cdot 4i = 20i^2 \]
03

Simplify the Result

Recall that \( i^2 = -1 \). Thus, we can rewrite \( 20i^2 \) as:\[ 20i^2 = 20(-1) = -20 \]Now, combine the simplified results:\[ 15i - 20 \]
04

Write in Standard Form

The product \( 15i - 20 \) can be written in standard form for complex numbers, which is \( a + bi \):\[ -20 + 15i \]

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Key Concepts

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

Distributive Property in Complex Numbers Multiplication
The distributive property is a fundamental rule in algebra that also applies to complex numbers. It states that a single term outside parentheses can be distributed to each term inside the parentheses. This means that you multiply the term outside by each term inside, one at a time.

In our original problem, this property is used to multiply the complex number by another expression. Here's a quick refresher on how it works:

  • Given two expressions, \( A(B + C) = A \cdot B + A \cdot C \)
  • This method ensures that every term in one expression is taken into account for multiplication with every term in the other expression.
  • In the problem \( 5i(3 + 4i) \), it means multiplying \( 5i \cdot 3 \) and then \( 5i \cdot 4i \) separately.
This methodical process sets the stage for further simplification in complex numbers by dealing with each term step-by-step.
Understanding the Imaginary Unit
The imaginary unit, represented by the symbol \( i \), is a core element of complex numbers. It is defined as the square root of \( -1 \), which means \( i^2 = -1 \). The use of \( i \) allows for the extension of numbers beyond the real number line, introducing what we know as complex numbers.

Here's why the imaginary unit is crucial:
  • It enables calculations and representations that are impossible with just real numbers alone.
  • In complex arithmetic, multiplying by \( i \) shifts the concept from real numbers to the imaginary plane.
For example, in the original solution, we see how \( 5i \cdot 4i = 20i^2 \) transforms into a real number, \(-20\), because \( i^2 \) equals \(-1\). Understanding this property is vital in dealing with products that involve imaginary parts.
Standard Form of Complex Numbers
The standard form of complex numbers is written as \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. This form is important because it clearly displays the real and the imaginary components.

Here's how the standard form works:
  • The \( a \) represents the real part of the complex number.
  • The \( b \) represents the coefficient of the imaginary part \( i \).
In the original exercise, after using the distributive property and simplifying the terms, the result \( 15i - 20 \) needs to be rearranged into the standard form, giving \( -20 + 15i \).

Writing in standard form helps in comprehension and ensures consistency when working with complex numbers in any mathematical operation.

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Most popular questions from this chapter

Show that \(x=a+b i\) is a solution to the equation \(x^{2}-2 a x+\left(a^{2}+b^{2}\right)=0\).

Use your graphing calculator in polar mode to generate a table for each equation using values of \(\theta\) that are multiples of \(15^{\circ}\). Sketch the graph of the equation using the values from your table.\(r=2 \sin 2 \theta\)

Graph each equation.\(\theta=\frac{\pi}{4}\)

Change each equation to rectangular coordinates and then graph.\(r=6 \cos \theta\)

Graph each equation.\(r=4 \sin 3 \theta\)
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