Problem 57 - Algebra 2

Question

REVIEW What is the product of the complex numbers \((5+i)\) and \((5-i) ?\) F 24 G 26 H \(25-i\) J \(26-10 i\)

Step-by-Step Solution

Answer

The product of \((5+i)\) and \((5-i)\) is 26.

Step 1: Understand the Problem

We need to find the product of two given complex numbers: \((5+i)\) and \((5-i)\). Then, we will have to match the result with one of the given options: F 24, G 26, H \(25-i\), J \(26-10 i\).

Step 2: Apply the Difference of Squares Formula

Notice that \((5+i)\) and \((5-i)\) are of the form \((a+b)(a-b)\) where \(a=5\) and \(b=i\). The difference of squares formula is \((a+b)(a-b) = a^2 - b^2\).

Step 3: Calculate the Squares

Calculate \(a^2\) and \(b^2\): \[a^2 = (5)^2 = 25\]\[b^2 = (i)^2 = -1\] (since \(i^2 = -1\))

Step 4: Find the Product

Substitute the calculated squares back to the difference of squares formula: \[(5+i)(5-i) = 5^2 - i^2 = 25 - (-1) = 25 + 1\]

Step 5: Simplify the Result

Simplifying gives: \[25 + 1 = 26\]. The result is a real number, not a complex number with an imaginary part.

Step 6: Verify the Solution

Verify the solution by checking against the provided options: F 24, G 26, H \(25-i\), J \(26-10 i\). Our answer matches option G 26.

Key Concepts

Difference of Squares

The difference of squares is a useful algebraic formula that appears often in mathematics, especially when working with polynomials and expressions involving complex numbers. The formula is expressed as \((a+b)(a-b) = a^2 - b^2\). This might look complicated at first glance, but let's break it down into simpler parts:

• "a" and "b" are any numbers or expressions: In this context, they can be real numbers, variables, or even complex numbers.
• "a+b" and "a-b" are factors: These are two binomials that multiply each other in the difference of squares.
• "a^2 - b^2" is the result: After using the formula, you get the difference between the square of "a" and the square of "b".

To see how this applies to complex numbers, consider \((5+i)(5-i)\). Here, \(5+i\) and \(5-i\) follow this setup where \(a = 5\) and \(b = i\). Plugging them into the formula, we get \(a^2 - b^2\), which simplifies the multiplication dramatically.

Imaginary Unit

The imaginary unit, often denoted as \(i\), is a fundamental concept in complex numbers. It is defined by the equation \(i^2 = -1\). Understanding the imaginary unit is crucial when working with complex numbers.

• Definition: At its core, \(i\) is defined such that \(i^2 = -1\). This is unlike any real number, because no real number squared would give you a negative result.
• How to Use \(i\): The unit itself makes calculations involving the square roots of negative numbers possible and is essential for expressing complex numbers.
• Notation: Complex numbers are expressed as \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.

In our example of multiplying \((5+i)\) by \(5-i\), we apply the property \(i^2 = -1\) during the calculation of \(b^2\), simplifying the expression \(-i^2\) to \(+1\). This demonstrates the practical use of the imaginary unit in simplifying calculations.

Multiplication of Complex Numbers

Multiplying complex numbers can seem challenging at first, but it follows straightforward rules. When multiplying two complex numbers, for instance, \((a+bi)(c+di)\), you'll distribute each term, just like when multiplying binomials in algebra.

• Distribute Each Term: Use the distributive property to expand. Multiply each term in the first complex number with each term in the second.
• Combine Like Terms: Adding and combining real and imaginary terms comes next. The real parts are added together and the imaginary parts are combined.
• Simplify: Ensure that all terms are simplified. Remember that \(i^2 = -1\), which helps in converting terms with \(i^2\) into real numbers.

In our solution, we used the difference of squares to simplify the multiplication of \((5+i)(5-i)\) directly to \((25 + 1)\), due to the property where \((-i^2 = 1)\). This example shows how properties of complex numbers like the imaginary unit make these operations more efficient and manageable.