Problem 66 - Intermediate Algebra

Question

Find the product of the given complex number and its conjugate. $$ 5+2 i $$

Step-by-Step Solution

Answer

The product is 29.

Step 1: Understand the Conjugate

The conjugate of a complex number \( a + bi \) is \( a - bi \). For the complex number \( 5 + 2i \), the conjugate is \( 5 - 2i \).

Step 2: Write the Expression for the Product

To find the product of a complex number and its conjugate, we write the expression: \((5 + 2i)(5 - 2i)\).

Step 3: Expand the Expression

Use the formula \((a + b)(a - b) = a^2 - b^2\). Here, \( a = 5\) and \( b = 2i \). Thus, the expression becomes \(5^2 - (2i)^2\).

Step 4: Calculate the Squares

Calculate \(5^2 = 25\) and \((2i)^2 = 4i^2\). Since \(i^2 = -1\), we have \(4i^2 = 4(-1) = -4\).

Step 5: Combine the Results

Substitute back into the expression: \(25 - (-4) = 25 + 4 = 29\).

Key Concepts

Conjugate of a Complex Number

A complex number is expressed in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part involving the imaginary unit \( i \). The conjugate of a complex number is essentially its twin without the imaginary part being positive. If the given complex number is \( 5 + 2i \), its conjugate would be \( 5 - 2i \). The only change here is the sign before the imaginary component.

• The conjugate essentially "flips" the sign of the imaginary part, changing it from \( +bi \) to \( -bi \).
• This flipping has a special significance when dealing with the multiplication of complex numbers as it can simplify many computations.

Understanding the concept of conjugate complex numbers is crucial because it helps in operations like division and finding the modulus or absolute value of a complex number. The product of a complex number and its conjugate always results in a real number, which is a key property often leveraged in different areas of mathematics.

Product of Complex Numbers

Multiplying complex numbers might seem tricky initially, but it's just like working with binomials. In general, for two complex numbers \( (a + bi) \) and \( (c + di) \), their product can be found using the distributive property:

• Multiply in the same way you would multiply two binomials using FOIL (First, Outer, Inner, Last).

Let's break down the steps using our specific complex number and its conjugate, \((5 + 2i)\) and \((5 - 2i)\). Note how we can use the difference of squares formula:

• We use \((5 + 2i)(5 - 2i) = 5^2 - (2i)^2\).
• This simplifies to calculating \( 25 - (2i)^2 \).
• Recognizing that \( (2i)^2 = 4i^2 \) and knowing that \( i^2 = -1 \), the equation further simplifies to \( 25 - 4(-1) \).

Ultimately, multiplying a complex number by its conjugate returns a real number, in this case, \( 29 \). This is because the imaginary components cancel each other out, leaving only the real component, which keeps calculations nice and tidy.

Imaginary Unit

The imaginary unit, denoted as \( i \), is a fundamental concept in complex numbers where \( i^2 = -1 \). While it's termed "imaginary," its usefulness is very real in mathematics, particularly when handling roots of negative numbers and in scientific calculations.

• The imaginary unit allows us to extend the real number system to complex numbers, helping us solve equations that do not have real solutions, like \( x^2 + 1 = 0 \).
• When you see \( i \), think of it as transforming an otherwise unsolvable problem into a feasible one.

In the context of our exercise, understanding \( i \) is essential because it turns a negative square into a helpful tool:

• When you calculate \( (2i)^2 \), you find it equals \( 4i^2 \), and because \( i^2 = -1 \), this transforms into \(-4\).

Overall, the imaginary unit \( i \) is indispensable in the realm of complex numbers, providing the gateway from purely real calculations to those involving both real and imaginary components.