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Chapter 2: Problem 10

Write the number as a pure imaginary number. $$\sqrt{-64}$$

Short Answer

Expert verified
The pure imaginary number for \( \sqrt{-64} \) is \( 8i \).

Step by step solution

01

Identify the negative number

In this case, the negative number is -64. The square root operation cannot be performed directly on negative numbers in the real number system.
02

Compute square root of the magnitude

Firstly, ignore the negative sign and compute the square root of the magnitude which is 64. The square root of 64 is 8.
03

Append the imaginary unit

Since the initial number was negative, we know that the result will be an imaginary number, so the final step is to append the imaginary unit \(i\) to the result of the square root computation. So, square root of -64 is \(8i\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Roots
When tackling square roots, it's essential to understand that we're determining which number, multiplied by itself, gives the original number. For example, the square root of 64 is 8 because 8 multiplied by 8 equals 64.
  • Square roots are denoted by the radical symbol \(\sqrt{}\).
  • They yield positive results, known as the principal square root.
However, when you encounter negative numbers under the radical, this challenges the regular square root operation. In standard mathematics, you cannot directly compute the square root of a negative number, simply because no real number squared will give a negative result. This is where the concept of imaginary numbers comes into play, specifically involving the imaginary unit, to handle such situations.
Negative Numbers
Negative numbers are those less than zero and are indicated by a minus sign \((-\)). They play a critical role across various mathematical contexts. But they have a unique impact when it comes to square roots.
  • Squaring any real number, whether positive or negative, results in a non-negative outcome.
  • No real number squared equals a negative number (like \(-64\)).
This reasoning leads to an important rule: in the real number system, square roots of negative numbers are undefined. Instead, we turn to imaginary numbers to accommodate this scenario. This involves introducing a new kind of unit that helps express these numbers, leading us directly into the fascinating realm of complex numbers.
Imaginary Unit
The imaginary unit is a mathematical concept created to overcome the limitations of square roots on negative numbers. It is denoted by \(i\) and is defined as the square root of \(-1\). This simple yet powerful idea expands the number system into the realm of complex numbers.
  • The equation \(i^2 = -1\) forms the foundation of all calculations involving imaginary numbers.
  • This allows expressions like \(\sqrt{-64}\) to be rewritten as \(\sqrt{64} \times \sqrt{-1} = 8i\).
By using the imaginary unit, each negative number under a radical can be represented with real square roots multiplied by \(i\). In this context, \(8i\) represents a pure imaginary number, providing a structured way to handle and understand operations that initially seem impossible.

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Most popular questions from this chapter

A security firm currently has 5000 customers and charges \(\$ 20\) per month to monitor each customer's home for intruders. A marketing survey indicates that for each dollar the monthly fee is decreased, the firm will pick up an additional 500 customers. Let \(R(x)\) represent the revenue generated by the security firm when the monthly charge is \(x\) dollars. Find the value of \(x\) that results in the maximum monthly revenue.

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=|x| ; g(x)=\frac{x}{x-3}$$

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=x-2 ; g(x)=2 x^{2}-x+3$$

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=\frac{-x+1}{2 x+3} ; g(x)=\frac{1}{x^{2}+1}$$

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