Problem 65 - Intermediate Algebra

Question

Find the product of the given complex number and its conjugate. $$ 2+6 i $$

Step-by-Step Solution

Answer

The product is 40.

Step 1: Define the Complex Conjugate

For any complex number, say \( a + bi \), its complex conjugate is \( a - bi \). Here, the given complex number is \( 2 + 6i \), so its conjugate is \( 2 - 6i \).

Step 2: Write the Expression for the Product

We need to find the product of the given complex number and its conjugate. This expression becomes \((2 + 6i)(2 - 6i)\).

Step 3: Apply the Difference of Squares Formula

Recall that \((a + b)(a - b) = a^2 - b^2\). In this case, \( a = 2 \) and \( b = 6i \), so the product is \(2^2 - (6i)^2\).

Step 4: Calculate Each Term

First, calculate \(2^2 = 4\). Then calculate \((6i)^2 = (6^2)(i^2) = 36(-1) = -36\).

Step 5: Simplify the Result

Substitute the calculated terms back into the expression: \(4 - (-36) = 4 + 36 = 40\).

Key Concepts

Complex Conjugate

The concept of a complex conjugate is an essential part of working with complex numbers. Each complex number can be written in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit, satisfying \( i^2 = -1 \). The complex conjugate of \( a + bi \) is \( a - bi \). This means to find the conjugate, you simply change the sign of the imaginary part. For example, the complex conjugate of \( 2 + 6i \) is \( 2 - 6i \).
Using the complex conjugate has several benefits, especially in simplifying calculations involving complex numbers. For instance, it allows us to rationalize denominators in division problems involving complex numbers or, as we'll see in this case, find products that simplify using other mathematical properties.

Difference of Squares

When dealing with the product of a complex number and its conjugate, the difference of squares formula is a powerful tool. The formula states: \[(a + b)(a - b) = a^2 - b^2\]In this context, we use it to simplify expressions like \((2+6i)(2-6i)\).

• Identify \( a \) and \( b \) in the expression: Here, \( a = 2 \) and \( b = 6i \).
• Apply the formula: You calculate \( a^2 \) and \( b^2 \) to simplify the expression.
  • \( a^2 = 2^2 = 4 \)
  • \( b^2 = (6i)^2 = 36i^2 = 36(-1) = -36 \)

Then, substitute back into the formula to get the final simplified result. This demonstrates why multiplying a complex number by its conjugate yields a real number, as the imaginary parts cancel out.

Algebraic Expressions

Understanding algebraic expressions is crucial to solve problems involving complex numbers effectively. Algebraic expressions consist of variables, constants, and operators put together to define a relationship or calculate a value. In working with complex numbers, you often manipulate algebraic expressions using standard algebraic techniques to simplify calculations. For instance, when multiplying expressions like \((2+6i)(2-6i)\), familiarity with expanding and simplifying expressions is required. Here’s how you apply it:- Multiply terms: Use the distributive property, also known as the FOIL method for binomials- Simplify using known identities: Recognize familiar patterns such as the difference of squares to make calculations straightforward- Substitute back into expressions: Make sure each term is correctly substituted back into the original expressionSimplifying algebraic expressions involving complex numbers requires practice to recognize and apply patterns and identities effectively.