Problem 47 - Algebra 2

Question

REVIEW If \(i=\sqrt{-1},\) then \(5 i(7 i)=\) \(\mathrm{F} 70\) \(\mathrm{G} 35\) \(\mathrm{H}-35\) \(\mathrm{J}-70\)

Step-by-Step Solution

Answer

Option H is correct: \(-35\).

Step 1: Understanding the Problem

The problem gives us the expression \(5i(7i)\) where \(i\) is the imaginary unit \(\sqrt{-1}\). We need to simplify this expression and find the correct answer from the given options.

Step 2: Applying Imaginary Unit Property

Recall that \(i = \sqrt{-1}\), thus \(i^2 = -1\). This property of imaginary numbers allows us to simplify expressions involving \(i\).

Step 3: Simplifying the Expression

First, multiply the two imaginary numbers: \(5i \times 7i = 35i^2\).

Step 4: Substituting the Value of \(i^2\)

Now, substitute \(i^2 = -1\) into the expression: \(35i^2 = 35(-1) = -35\).

Step 5: Selecting the Correct Option

Compare the simplified result \(-35\) with the given options. The correct answer is the one matching \(-35\), which is \(\text{H}\).

Key Concepts

Imaginary Unit

The imaginary unit, denoted as \(i\), is a fundamental concept in complex numbers. Simply put, \(i\) is defined such that \(i = \sqrt{-1}\). This definition is crucial because it allows mathematicians and scientists to work with square roots of negative numbers, which are not possible in the realm of real numbers. The existence of \(i\) expands the number system into what we call complex numbers.

Complex numbers have the form \(a + bi\), where \(a\) and \(b\) are real numbers. Here, \(a\) is the real part, and \(bi\) is the imaginary part. Understanding the imaginary unit helps in simplifying and calculating expressions involving complex numbers effectively.

Multiplying Complex Numbers

When multiplying complex numbers, the distributive property is used, and it involves treating \(i\) as a variable at first. Suppose you multiply two simple complex numbers: \((a + bi)(c + di)\).

• First, multiply as you would with binomials: \(ac + adi + bci + bdi^2\).
• Combine like terms. Don't forget the key property, \(i^2 = -1\).

In our specific problem \(5i \times 7i\), since both terms are purely imaginary, you handle it similarly:

• Multiply the coefficients: \(5 \times 7 = 35\).
• Multiply \(i \times i = i^2\).

Remembering the property of \(i\) helps us to then substitute \(i^2\) with \(-1\). This transforms the expression into real numbers.

Properties of i

The properties of the imaginary unit \(i\) are surprisingly straightforward. The primary property that is always used is that \(i^2 = -1\).

Understanding powers of \(i\) is helpful:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\) (since \(i^3 = i^2 \cdot i = -1 \cdot i = -i\))
• \(i^4 = 1\) (since \(i^4 = (i^2)^2 = (-1)^2 = 1\))

Due to this cycle of powers repeating every four, you can easily simplify expressions containing powers of \(i\) beyond \(i^2\). For example, \(i^5 = i\), as it resumes the cycle starting with \(i\). This cyclic property is a powerful tool for complex number calculations.

Simplifying Expressions with i

Simplifying expressions with \(i\) requires both algebraic manipulations and applying the properties of \(i\). The key steps are:

• Multiply the coefficients and separate the terms involving \(i\).
• Use the property \(i^2 = -1\) to simplify any \(i^2\) in your expression.
• Combine like terms and simplify further if needed.

In our exercise, by multiplying \(5i (7i)\), we started with \(35i^2\). Using the property \(i^2 = -1\), we replaced \(i^2\) with \(-1\), leading us to \(35(-1) = -35\).

This step-by-step methodical approach ensures that expressions with \(i\) are reduced to their simplest form, turning potentially complex calculations into straightforward arithmetic.