Problem 14 - Algebra and Trigonometry with Analytic Geometry

Question

Write the expression in the form \(a+b i\), where \(a\) and \(b\) are real numbers. $$ (4+9 i)(4-9 i) $$

Step-by-Step Solution

Answer

The expression simplifies to 97, a real number.

Step 1: Apply the Difference of Squares Formula

Recall that for any two numbers, \((x+y)(x-y) = x^2 - y^2\). Here, let's identify \(x = 4\) and \(y = 9i\). You will use this formula to simplify the expression.

Step 2: Substitute and Simplify

Substitute \(x = 4\) and \(y = 9i\) into the formula: \((4+9i)(4-9i) = 4^2 - (9i)^2\). Simplifying within the parentheses gives us \(4^2 = 16\) and \((9i)^2 = 81i^2\).

Step 3: Substitute i² with -1

Recall that \(i^2 = -1\). Therefore, \(81i^2 = 81(-1) = -81\). Now substitute this back to get the expression: \(16 - (-81)\).

Step 4: Simplify the Expression

Perform the subtraction: \(16 - (-81) = 16 + 81\). Therefore, the expression simplifies to \(97\), which is a real number with no imaginary part.

Key Concepts

Difference of Squares

The difference of squares is a valuable concept in algebra when dealing with expressions such as \(x^2 - y^2\). It provides a simple method for multiplying two conjugates, like \( (x+y)(x-y) \), to achieve a quick simplification while avoiding complex multiplication.

• This formula can be identified as \( (x+y)(x-y) = x^2 - y^2 \).
• Instead of expanding every term, it allows you to directly find the simplified result.

In the given problem, we used this formula by identifying \(x = 4\) and \(y = 9i\). This substituted into the formula results in \(4^2 - (9i)^2\), leading us swiftly to the simpler terms without extensive calculations. This method shows why difference of squares is an efficient shortcut for expressions of this type.

Imaginary Unit

The imaginary unit, represented by \( i \), is a fundamental concept in complex numbers. It is defined by the property \(i^2 = -1 \). Understanding the imaginary unit helps in handling complex numbers, which are numbers that include a real part and an imaginary part.

• If you square any real number, the result is always non-negative, but the square of \(i\) is \(-1\).
• In our exercise, \(9i\) is an imaginary number, and squaring it results in \(81i^2\), which we know transforms into \(-81\) because \(i^2 = -1\).

Through this transformation, imaginary numbers become central in expressing and manipulating complex number expressions. They allow for problems involving complex conjugates, like our example, to be solved neatly.

Real Numbers

Real numbers are the backbone of most mathematical calculations. They include all the numbers on the number line, encompassing both positive and negative numbers, as well as zero. In the context of complex numbers, the real part is often combined with an imaginary part to form a complex number.

• Real numbers do not involve the imaginary unit \(i\).
• In our solution, after applying the difference of squares and simplifying the imaginary component, we end up with the real number \(97\).

This final number, \(97\), shows that sometimes, after performing operations on complex numbers, the result can be purely real, with no remaining imaginary part. This exemplifies how complex number operations can result in a range of outcomes, including real numbers. It underscores the flexibility and broad applicability of complex numbers in solving problems across mathematics.