Problem 33 - A Graphical Approach to Precalculus with Limits

Question

Multiply or divide as indicated. Simplify each answer. $$\sqrt{-13} \cdot \sqrt{-13}$$

Step-by-Step Solution

Answer

The simplified answer is -13.

Step 1: Write the expression using imaginary numbers

The expression involves the square root of negative numbers. Using the property that \(\sqrt{-a} = i\sqrt{a}\), where \(i\) is the imaginary unit, rewrite each square root term. Thus, \(\sqrt{-13} = i\sqrt{13}\). So, the expression becomes \(\sqrt{-13} \cdot \sqrt{-13} = (i\sqrt{13})(i\sqrt{13})\).

Step 2: Apply the properties of imaginary numbers

Apply the property of multiplication for imaginary numbers: \((i\sqrt{13})(i\sqrt{13}) = i^2 (\sqrt{13})^2\). Since \(i^2 = -1\), the expression becomes \(-1 \cdot 13\).

Step 3: Simplify the expression

Now, calculate the expression: \(-1 \cdot 13 = -13\). This is the simplified form of your expression.

Key Concepts

Square Roots

The concept of square roots often begins with the number we can multiply by itself to produce a given non-negative number. However, when dealing with negative numbers under the square root symbol, we encounter square roots of negative numbers. Usually, the square root operation expects non-negative values, because no real number squared results in a negative number.

For example, \(\sqrt{4}\) equals 2, because 2 multiplied by itself results in 4. But what about \(\sqrt{-4}\)? This is where imaginary numbers come into play. To deal with such cases, mathematicians introduced the imaginary unit, denoted as \(i\), to represent the square root of \(-1\). Hence, \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i\).

• Real square roots produce real numbers.
• Square roots of negative numbers lead us to imaginary numbers.

Understanding the distinction between real and imaginary numbers is crucial when working with the square roots of negative numbers. Imaginary numbers allow mathematical operations with a broader spectrum of numbers beyond the visible number line.

Imaginary Unit

The imaginary unit 'i' is a fascinating concept in mathematics that allows us to expand the realm of numbers from real to complex. It is defined as the square root of negative one, meaning \(i^2 = -1\). The introduction of 'i' addresses the limitations when sqaring numbers and encountering a negative result.

The imaginary unit is a key building block in forming complex numbers, which are expressions composed of a real part and an imaginary part. A generic complex number is expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers.

• 'i' enables the square root of negative numbers.
• The complexity of numbers increases using 'i'.

Employing the imaginary unit allows one to solve equations that otherwise don't have real solutions, such as quadratic functions with no x-axis intercepts. Mastering the use of 'i' helps in unlocking different mathematical concepts and functions.

Multiplication of Complex Numbers

Multiplying complex numbers, which involve both real and imaginary parts, uses the distributive property much like multiplying binomials. When multiplying, each part of one complex number gets multiplied by each part of the other.

Consider two complex numbers, \(a + bi\) and \(c + di\). To multiply, follow these steps:

• Multiply real parts: \(a \cdot c\)
• Multiply real by imaginary: \(a \cdot di\) and \(bi \cdot c\)
• Multiply imaginary parts: \(bi \cdot di = bdi^2\)

The crucial step is handling \(i^2\), since \(i^2\) equates to \(-1\). For example, multiplying \(i\sqrt{13}\) by itself follows this principle:

\[ (i\sqrt{13})(i\sqrt{13}) = i^2 (\sqrt{13})^2 \]

After recognizing \(i^2 = -1\), the expression simplifies to \(-1 \cdot 13\), resulting in \(-13\).

• Distribute terms like binomials.
• Apply the identity \(i^2 = -1\) for simplification.

This method ensures a proper understanding of how to handle these interactions and simplifications within complex multiplication.