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Chapter 3: Problem 65

Multiply as indicated. Write each product in standard form. $$(\sqrt{6}+i)(\sqrt{6}-i)$$

Short Answer

Expert verified
The product is 7.

Step by step solution

01

Identify the Structure

The expression \((\sqrt{6} + i)(\sqrt{6} - i)\) is in the form of a conjugate pair, \((a + bi)(a - bi)\), where \(a = \sqrt{6}\) and \(b = 1\). This is a special product called the difference of squares.
02

Apply the Difference of Squares Formula

The difference of squares formula is \((a + b)(a - b) = a^2 - b^2\). For complex numbers, when multiplying a conjugate pair, the result is \(a^2 + b^2\) when considering \(i^2 = -1\).
03

Calculate \(a^2\)

Calculate \(a^2\) where \(a = \sqrt{6}\). So, \(a^2 = (\sqrt{6})^2 = 6\).
04

Calculate \(b^2\)

Calculate \(b^2\) where \(b = 1\). So, \(b^2 = (1)^2 = 1\).
05

Combine Results

According to the modified difference of squares for complex numbers, the product is \(a^2 + b^2\). Substitute \(a^2 = 6\) and \(b^2 = 1\) to get the result: \(6 + 1 = 7\).

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Key Concepts

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

Conjugate Pairs
Complex numbers often come in pairs, known as conjugate pairs. When a complex number is expressed in the form \(a + bi\), its conjugate pair is \(a - bi\). These pairs have a special property that makes them very interesting.
  • When multiplied together, the imaginary components \((bi)\) cancel each other out.
  • This results in a simplified expression that is purely real.
In our exercise, the expression \((\sqrt{6} + i)(\sqrt{6} - i)\) is a perfect example. Here, \(a = \sqrt{6}\) and \(b = 1\). When we multiply these conjugates, their imaginary parts cancel out nicely. You'll find that the calculation becomes much simpler, focusing only on the real parts.
Difference of Squares
The difference of squares is a powerful algebraic identity, expressed as \((a + b)(a - b) = a^2 - b^2\). However, when dealing with complex numbers, there is a slight twist.
  • For complex conjugate pairs \((a + bi)(a - bi)\), the identity modifies to \(a^2 + b^2\) because \(i^2 = -1\).
  • This means the imaginary contribution becomes positive when squaring the imaginary part.
Applying this knowledge to our example, we calculated \(\sqrt{6}^2\) which is \(6\), and \(1^2\) is \(1\). Adding these gives us \(a^2 + b^2 = 6 + 1 = 7\). These steps demonstrate the elegance and utility of using the difference of squares with conjugate pairs.
Standard Form
The standard form for a complex number is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers. This is the typical way to present complex numbers, making it easy to understand both their real and imaginary components.
  • Even after performing operations like multiplication, results are typically returned to standard form.
  • This ensures clarity and uniformity in the way complex numbers are reported.
In our exercise, the final product of the conjugate pair multiplication resulted in the real number \(7\). Since there is no imaginary component, it simplifies neatly and is already in standard form. This makes it clear that the multiplication of the specific conjugate pairs yielded a purely real result, free of an imaginary part.

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Most popular questions from this chapter

Find all rational zeros of each polynomial function. $$P(x)=\frac{10}{7} x^{4}-x^{3}-7 x^{2}+5 x-\frac{5}{7}$$

Solve each formula for the indicated variable. Leave \(\pm\) in answers when appropriate. Assume that no denominators are \(0 .\) $$\mathscr{A}=\pi r^{2} \quad \text { for } r$$

Find all rational zeros of each polynomial function. $$P(x)=\frac{1}{6} x^{4}-\frac{11}{12} x^{3}+\frac{7}{6} x^{2}-\frac{11}{12} x+1$$

Solve each formula for the indicated variable. Leave \(\pm\) in answers when appropriate. Assume that no denominators are \(0 .\) $$F=\frac{k M v^{4}}{r} \text { for } v$$

For each polynomial, at least one zero is given. Find all others analytically. $$P(x)=x^{4}-41 x^{2}+180 ; \quad-6 \text { and } 6$$
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