Problem 48 - College Algebra Essentials

Question

In Exercises \(45-50,\) perform the indicated operation(s) and write the result in standard form. $$ (4-i)^{2}-(1+2 i)^{2} $$

Step-by-Step Solution

Answer

The solution to the complex number arithmetic problem (4-i)^{2}-(1+2 i)^{2} is 18 + 4i.

Step 1 Calculate the squares

First square both parts of the equation separately, using the formula \((a − b)^2 = a^2 − 2ab + b^2\) and \((a + b)^2 = a^2 + 2ab + b^2\). This will give you \((4-i)^{2} = 4^2 - 2*4*(-i) + (-i)^2\) and \((1+2i)^{2} = 1^2 + 2*1*2i + (2i)^2\).

Step 2 Perform the calculation

Perform the calculations to find the answers for the squares. For \((4-i)^{2} = 16 + 8i - 1\), and for \((1+2i)^{2} = 1 + 4i - 4\). Remember that \(i^2 = -1\).

Step 3 Simplify both squares

Now you can simplify the squared results. This gives you \((4-i)^{2} = 15 + 8i\) and \((1+2i)^2 = -3 + 4i\).

Step 4 Subtract the second square from the first

Subtract the second square from the first, \( (15 + 8i) - (-3 + 4i) \).

Step 5 Perform the subtraction

Perform the subtraction, which yields the result \( 15 - (-3) + 8i - 4i = 18 + 4i \). Thus, the solution in standard form for the complex number is \(18 + 4i\).

Key Concepts

Complex Number Algebra

Understanding the algebra of complex numbers can first seem daunting, but it becomes much easier once you grasp a few basic principles. Complex numbers consist of two parts: a real part and an imaginary part. They are typically written in the standard form of \( a + bi \), where \( a \) is the real component, \( b \) is the imaginary coefficient, and \( i \) is the imaginary unit with the property that \( i^2 = -1 \).

When performing operations with complex numbers, such as addition, subtraction, and multiplication, you combine the real parts and the imaginary parts separately, always keeping in mind the special property of \( i \). For example, when you add \( 3 + 2i \) and \( 1 + 4i \), you add the real parts to get 4, and the imaginary parts to get \( 6i \), resulting in a sum of \( 4 + 6i \).

Subtraction works similarly, while for multiplication, the distributive property is applied, similar to how you would expand binomials. One of the key aspects to remember is that any time you encounter \( i^2 \), it should be replaced with \( -1 \), simplifying the expression further. This aspect is crucial when squaring complex numbers, as seen in the textbook exercise.

Simplifying Complex Expressions

To simplify complex expressions, one must carefully combine like terms and apply algebraic rules diligently. In the context of complex numbers, this process typically involves adding or subtracting real parts with real parts and imaginary parts with imaginary parts separately. It is essential when simplifying to use the key property that \( i^2 = -1 \), which helps in eliminating the \( i^2 \) terms and reduces the expression to its simplest form.

For instance, when confronted with the expression \( (15 + 8i) - (-3 + 4i) \), simplify it by subtracting the real part, which gives \( 15 - (-3) \), and the imaginary part, which gives \( 8i - 4i \). After calculating, you obtain \( 18 + 4i \), which has no like terms left to combine, thus presenting the simplified standard form of the complex number. This simplified form is not just the answer but also an easier version for further calculations or analysis, if necessary.

Square of a Binomial

The square of a binomial is a foundational concept in algebra that also applies to complex numbers. A binomial is an expression consisting of two terms. When you square a binomial, you are essentially multiplying the binomial by itself. The pattern that emerges is represented by the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), where \( a \) and \( b \) are the terms of the binomial.

In the context of complex numbers, when squaring a binomial such as \( (4 - i) \), it's important to apply this formula carefully, knowing that the term that involves \( i \) will need further simplification due to the property of \( i^2 \). Thus, \( (4 - i)^2 \) becomes \( 4^2 - 2 \cdot 4 \cdot (-i) + (-i)^2 \), which simplifies to \( 16 + 8i - 1 \), because \( i^2 \) is equal to \( -1 \). The final simplified form, taking the square of a binomial into account along with the properties of \( i \), is essential for complex number algebra and is illustrated in the textbook example solution.