Problem 91 - Calculus for Biology and Medicine

Question

Simplify each expression and write it in the standard form \(a+b i\). \((6-i)(6+i)\)

Step-by-Step Solution

Answer

The expression simplifies to \(37 + 0i\).

Step 1: Apply the difference of squares formula

Recognize that the expression \((6-i)(6+i)\) is a difference of squares. The formula for the difference of squares is \((a-b)(a+b) = a^2 - b^2\). Here, \(a = 6\) and \(b = i\).

Step 2: Calculate each square

Calculate \(a^2\) and \(b^2\). For \(a^2\), we have \(6^2 = 36\). For \(b^2\), since \(b = i\), \(b^2 = i^2 = -1\).

Step 3: Substitute into the formula

Substitute \(a^2\) and \(b^2\) into the difference of squares formula: \(a^2 - b^2 = 36 - (-1)\).

Step 4: Simplify the expression

Simplify the expression, taking care to correctly handle the double negative: \(36 - (-1) = 36 + 1 = 37\).

Step 5: Write in standard form

The simplified expression is \(37\), which can be written in the form \(a + bi\) as \(37 + 0i\). This is already in standard form, where \(a = 37\) and \(b = 0\).

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic concept that simplifies the multiplication of two binomials. When you see an expression like

• the product of the difference and sum of two terms
• look for an opportunity to use the formula:
• \((a - b)(a + b) = a^2 - b^2\).

This formula pulls apart the structure of a problem, making it simpler. In the given exercise,

• the expression
• you were asked to simplify is
• \((6-i)(6+i)\), which fits perfectly into this formula.
• Here, the values are \(a = 6\) and \(b = i\), demonstrating
• how quickly a complex multiplication challenge can turn into the easier task of calculating two squares and their difference.

By rearranging the expression into a difference of squares, you are able to compute each part separately and efficiently.This transform reduces complexity, turning a hard-looking problem into quick arithmetic.

Imaginary Unit

At the core of complex numbers lies the imaginary unit, denoted as \(i\). Understanding this unit is vital when dealing with expressions involving complex numbers, like those in the exercise. The imaginary unit \(i\) is defined by the property that

• \(i^2 = -1\).

In other words, \(i\) serves as a number whose square is negative, stepping outside the realm of real numbers and enabling computations with square roots of negative numbers. To see this in the exercise, note how the expression uses \(b = i\).

• When you calculate \(b^2\), it simplifies to
• you're using the rule \(i^2 = -1\). Knowing this property simplifies calculating squares with imaginary numbers
• and allows the solution process to move smoothly.

Embrace the concept of \(i\), as it expands traditional arithmetic into new areas, making calculations like these not just possible, but straightforward.

Complex Conjugates

Complex numbers often appear alongside their counterparts, known as complex conjugates. A complex number, typically represented as \(a + bi\), has a conjugate expressed as \(a - bi\). Such pairs are intriguing because they have beautiful mathematical properties, particularly when multiplied.In the exercise at hand, notice how

• the structure of
• \((6-i)(6+i)\) inherently involves a complex number and its conjugate.
• This special product simplifies directly to a real number
• without imaginary parts.
• This occurs because the imaginary parts cancel out,
• due to the nature of the square of \(i\) and the subtraction formula.

Thus, the expression initially appears complicated, but the use of conjugates converts it back into the real number system, resulting in

• a solution
• like
• \(37\)

— much simpler looking compared to where it started. This highlights the elegance and efficiency of working with complex conjugates.