Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-34: Write the expression in the form \(a+b i\), where \(a\) and \(b\) are real numbers. $$ (3+4 i)(3-4 i) $$

Step-by-Step Solution

Answer

The expression simplifies to 25 + 0i.

Step 1: Understand the Identity

Recognize that the expression \((3+4i)(3-4i)\) follows the pattern of the difference of squares, which is \((a+bi)(a-bi) = a^2 + b^2\). This identity simplifies the multiplication of complex conjugates.

Step 2: Identify the Parts

Here, the real part \(a\) is 3 and the imaginary part \(b\) is 4. Notice that \((3+4i)\) and \((3-4i)\) are complex conjugates.

Step 3: Apply the Formula

Substitute the values into the difference of squares formula: \[(3+4i)(3-4i) = 3^2 + 4^2 = 9 + 16\]

Step 4: Simplify the Result

Add together the results from the previous calculation: \[9 + 16 = 25\]

Step 5: Write in the Form a + bi

Since \(25\) is a real number, we can write this in the form \(a + bi\) where \(a = 25\) and \(b = 0\), so the expression is simply \(25 + 0i\).

Key Concepts

Complex Conjugates

Complex conjugates play a crucial role in simplifying complex number expressions. When dealing with complex numbers, especially in multiplication, using complex conjugates can simplify calculations immensely. A complex number is typically expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit. A complex conjugate takes the form \(a - bi\).

• The original complex number: \(3 + 4i\)
• The conjugate: \(3 - 4i\)

Recognizing these pairs is essential. When you multiply a complex number by its conjugate, the imaginary parts cancel each other out, resulting in a real number. This unique property makes complex conjugates extremely useful in complex number arithmetic.In the given exercise, both numbers \((3+4i)\) and \((3-4i)\) are complex conjugates, and their multiplication uses this concept to achieve a real number as a result.

Difference of Squares

The difference of squares is an algebraic identity that states \((a+b)(a-b) = a^2 - b^2\). This principle extends to complex numbers, becoming \((a+bi)(a-bi) = a^2 + b^2\). This adaptation helps simplify products of complex conjugates.Here is how it works step by step:

• Identify \(a\) and \(b\) from the complex numbers. In our exercise, \(a = 3\) and \(b = 4\).
• Plug these values into the difference of squares formula for complex numbers: \(a^2 + b^2 = 3^2 + 4^2\)
• Calculate \(a^2\) and \(b^2\):
• \(3^2 = 9\)
• \(4^2 = 16\)
• Add these results: \(9 + 16 = 25\)

The result of this calculation is 25, a real number. This illustrates how the difference of squares makes it easier to multiply complex conjugates.

Imaginary Numbers

Imaginary numbers are fundamental to complex numbers and involve the unit \(i\), which is defined as the square root of \(-1\). Real numbers are expanded by incorporating \(i\), forming complex numbers.

• The concept: \(i = \sqrt{-1}\).
• Allows real numbers to extend into the complex plane.
• Expressed as: \(a + bi\) where \(a\) is the real part and \(bi\) is the imaginary part.

In the given exercise, \(4i\) represents the imaginary part of the complex numbers \(3 + 4i\) and \(3 - 4i\). When these complex numbers are multiplied, the distinctive property of \(i\) (i.e., \(i^2 = -1\)) helps ensure that the imaginary components negate each other, leaving a real number.This creative fusion of real and imaginary numbers leads to the fascinating world of complex numbers, where both elements can be utilized in practical applications across mathematics, engineering, and physics.