Problem 61 - Precalculus Student Solutions Manual 5th

Question

Find each product. Write the answer in standard form. $$(\sqrt{6}+i)(\sqrt{6}-i)$$

Step-by-Step Solution

Answer

The product is \(5\).

Step 1: Identify the factors

In the given problem, the factors are \(\text{(\textbackslash sqrt}6 + i)\) and \(\text{(\textbackslash sqrt}6 - i)\).

Step 2: Use the conjugate product rule

The factors are conjugates of each other \((a + bi)(a - bi) = a^2 - b^2i^2\), where \(a = \text{\textbackslash sqrt}6\) and \(b = 1\).

Step 3: Substitute values into the formula

Substitute \(a\) and \(b\) into the formula: \((\textbackslash sqrt}6)^2 - (1)^2 \text{-}i^2.\)

Step 4: Perform the arithmetic operations

Calculate \( (\textbackslash sqrt}6\text{)^2 \) = 6, \( 1^2 = 1\), and \((-i^2 = -(-1) = 1).\)

Step 5: Compute the product

Combine the results from Step 4: \( 6 \text{-}1= 5.)\)

Key Concepts

Conjugate Product Rule

In this exercise, we are tasked with multiplying two complex numbers. The key to solving this particular problem is understanding and using the conjugate product rule.
When dealing with complex numbers, a conjugate is created by changing the sign of the imaginary part. For example, the conjugate of \(\text{a} + i\text{b}\) is \(\text{a} - i\text{b}\).
The conjugate product rule states that when you multiply a complex number by its conjugate, the result is always a real number. The formula for this is: \((a + bi)(a - bi) = a^2 - b^2i^2\).
In this exercise, we are given \((\text{\textbackslash sqrt}6 + i)\text{ and }(\text{\textbackslash sqrt}6 - i)\) as factors. Notice that they are conjugates of each other, so we can apply the conjugate product rule.
By substituting \(a = \text{\textbackslash sqrt}6\) and \(b = 1\), and using the formula, we can simplify the multiplication process greatly. Later, the calculations involve basic arithmetic operations.

Imaginary Unit

The imaginary unit is a fundamental concept when dealing with complex numbers. It is denoted by \(i\) and has the property that \(i^2 = -1\).
This unique property allows us to extend the real number system to the complex number system.
In our exercise, the imaginary unit \(i\) is present in the factors \((\text{\textbackslash sqrt}6 + i)\text{ and }(\text{\textbackslash sqrt}6 - i)\).
When we use the conjugate product rule, we encounter \(i^2\). Substituting \(i^2\) with \(-1\) turns complex multiplications into simpler arithmetic problems.
For instance, when computing \(b^2i^2\), in our case \(b = 1\), we calculate \(1^2 \text{-}i^2 = 1 \text{-} (-1) = 1 + 1 = 2\).
Always remember, every time you see \(i^2\), replace it with \(-1\). This step reduces complexity and helps in solving the problem accurately.

Standard Form

The standard form of a complex number is written as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.
This form makes complex numbers easy to understand and manipulate.
In our exercise, we start with two complex factors \(\text{\textbackslash sqrt}6 + i\) and \(\text{\textbackslash sqrt}6 - i\).
After applying the conjugate product rule and performing the arithmetic, we need to ensure our final answer is in standard form.
The steps in the solution lead us to compute the product, starting from \(a^2 - b^2i^2\) and simplifying to get a real number.
In our case, the answer we get is 5, which is already in standard form because it is a real number without an imaginary part.
Always double-check to ensure that the final answer is properly formatted in standard form, especially if both real and imaginary parts are present.