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Chapter 8: Problem 59

Multiply. $$ (1+i)^{2}(1-i)^{2} $$

Short Answer

Expert verified
The result is \( 4 \).

Step by step solution

01

- Expand the Squares

First, expand \( (1+i)^{2} \) and \( (1-i)^{2} \) separately. Use the formula for squaring a binomial, \( (a+b)^2 = a^2 + 2ab + b^2 \).For \( (1+i)^{2} \):\[(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i + i^2\]Since \( i^2 = -1 \):\[1 + 2i + (-1) = 2i\]For \( (1-i)^{2} \):\[(1-i)^2 = 1^2 + 2(1)(-i) + (-i)^2 = 1 - 2i + i^2\]Since \( i^2 = -1 \):\[1 - 2i + (-1) = -2i\]
02

- Multiply the Results

Now that we have expanded the squares, we multiply them together:\[2i \times -2i = -4i^2\]Since \( i^2 = -1 \):\[-4(-1) = 4\]

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Key Concepts

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

complex numbers
Complex numbers are numbers that have both a real part and an imaginary part. The standard form of a complex number is written as:
\[a + bi\]
where:
  • \(a\) is the real part
  • \(bi\) is the imaginary part
The imaginary part involves the imaginary unit, which is represented by \(i\) and is defined as \(i^2 = -1\). Using complex numbers helps solve equations that don't have real solutions. They are useful in fields like engineering and physics.
binomial expansion
Binomial expansion involves expanding an expression that is raised to a power. In our problem, we needed to expand \((1+i)^2\) and \((1-i)^2\). For this, we use the formula for the square of a binomial:
\[ (a+b)^2 = a^2 + 2ab + b^2 \]
Let's look at how this applies to our problem:
  • For \((1+i)^2\):
    \[ (1+i)^2 = 1^2 + 2(1)(i) + i^2 \]
    Substituting \(i^2\) with -1:
    \[ 1 + 2i - 1 = 2i \]
  • For \((1-i)^2\):
    \[ (1-i)^2 = 1^2 + 2(1)(-i) + (-i)^2 \]
    Replacing \(i^2\) with -1:
    \[ 1 - 2i - 1 = -2i \]
This way, we transformed complex binomials into simpler forms.
imaginary unit i
The imaginary unit, denoted as \(i\), is a fundamental concept in the realm of complex numbers. Its core property is that:
\[ i^2 = -1 \]
This means that \(i\) stands for the square root of -1. It doesn't exist on the real number line but is useful in solving mathematical problems that involve the square roots of negative numbers. For example, in our exercise, recognizing that \(i^2 = -1\) allows us to simplify terms involving \(i^2\). Converting \(i^2\) into -1 made the calculations clearer and simpler.
algebraic multiplication
Algebraic multiplication is the process of multiplying algebraic expressions. In our problem, this involves multiplying the results of expanded binomials. After expanding \((1+i)^2\) and \((1-i)^2\), we multiplied the simplified forms:
\[2i \times -2i = -4i^2 \]
Through substitution, knowing \(i^2 = -1\):
\[-4(-1) = 4\]
This way, the use of algebraic multiplication, along with the properties of \(i\), led to a final real number result. Breaking it down step-by-step helps understand how multiplying complex numbers brings about a real number solution, demonstrating the elegance of algebraic operations with complex numbers.

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Most popular questions from this chapter

Find each quotient. $$ \frac{-1+5 i}{3+2 i} $$

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