Problem 73 - Algebra and Trigonometry

Question

Recall that the symbol \(\overline{z}\) represents the complex con- jugate of \(z .\) If \(z=a+b i\) and \(w=c+d i,\) prove each statement. \(z \cdot \overline{z}\) is a real number

Step-by-Step Solution

Answer

The product \( z \cdot \overline{z} = a^2 + b^2 \), a real number.

Step 1: Understand the Complex Conjugate

The complex conjugate of a complex number \( z = a + bi \) is \( \overline{z} = a - bi \). The conjugate is found by changing the sign of the imaginary part.

Step 2: Multiply the Complex Number by its Conjugate

Multiply \( z \) and \( \overline{z} \):\[ z \cdot \overline{z} = (a + bi)(a - bi) \]

Step 3: Apply the Difference of Squares Formula

Recognize that the multiplication \( (a + bi)(a - bi) \) fits the formula for the difference of squares, \( (x + y)(x - y) = x^2 - y^2 \). For this problem:\[ a^2 - (bi)^2 = a^2 - (b^2)(i^2) \]

Step 4: Simplify Using Imaginary Unit Properties

Simplify using the property \( i^2 = -1 \):\[ a^2 - b^2(-1) = a^2 + b^2 \]

Step 5: Verify the Expression is Real

Notice that \( a^2 + b^2 \) consists only of real numbers, with no imaginary part. Therefore, the product \( z \cdot \overline{z} \) is a real number.

Key Concepts

Complex Number Multiplication

Complex numbers are fascinating as they blend real and imaginary numbers. When multiplying complex numbers, follow certain rules to ensure accuracy. Let's dive into how this works using a specific example: multiplying a complex number by its conjugate.

Consider the complex number, \( z = a + bi \). Its conjugate is \( \overline{z} = a - bi \). When multiplying these two, you employ the distributive property just as you do with real numbers. Here's how it looks:

• First, multiply \( a \) with \( a \), resulting in \( a^2 \).
• Then, multiply \( a \) with \(-bi\), giving \(-abi\).
• Next, multiply \( bi \) with \( a \), which equals \( abi \).
• Finally, multiply \( bi \) with \(-bi\), leading to \(-b^2i^2\).

After multiplicating, you'll notice that the imaginary components \(-abi\) and \( abi \) cancel each other out. What you're left with is \( a^2 - b^2i^2 \). This expression is crucial in understanding how complex multiplication works and setting the stage for further simplification.

Difference of Squares

The product \((a + bi)(a - bi)\) is a classic example of the difference of squares, a mathematical pattern that simplifies calculations. This method can transform potentially complex problems into solvable ones.

In our equation, \((a + bi)(a - bi)\), treat \( a \) as \( x \) and \( bi \) as \( y \) to align with the difference of squares formula: \( x^2 - y^2 \). Applying this to the equation gives you \( a^2 - (bi)^2 \).

This pattern effectively deals with squared terms, allowing you to break down expressions into manageable components. The key takeaway here is recognizing situations where you can apply the difference of squares to simplify and solve complex number operations.

Imaginary Unit Properties

The imaginary unit, \( i \), is defined by the property \( i^2 = -1 \). Understanding this is central to many calculations involving complex numbers. Let's see how this property simplifies expressions.

Returning to our problem, after applying the difference of squares formula, you get \( a^2 - b^2i^2 \). Brushing up your knowledge on the imaginary unit, replace \( i^2 \) with \(-1\). The expression therefore becomes \( a^2 - b^2(-1) \) or \( a^2 + b^2 \).

Notice this transformation eliminates the imaginary part, leaving us with a real number sum \( a^2 + b^2 \). Recognizing the property of \( i \) and applying it allows for the simplification of otherwise complex expressions. This understanding is key when working with polynomials and algebraic expressions involving imaginary numbers.