Chapter 3: Problem 46
Use a calculator to help answer the questions. Evaluate \((1+i)^{k}-(1-i)^{k}\) for \(k=4,8,\) and \(12 .\) Predict the value for \(k=16\)
Short Answer
Expert verified
For all given \(k\), the expression evaluates to 0, so \(k=16\) also predicts 0.
Step by step solution
01
Recognize the Structure
The expression \((1+i)^{k}-(1-i)^{k}\) involves complex numbers and powers. Here, \(i\) is the imaginary unit where \(i^2 = -1\). To solve for specific values of \(k\), we'll be evaluating this expression for each value of \(k\) given.
02
Calculate \( (1+i)^{4} \) and \( (1-i)^{4} \)
Use the binomial theorem or a calculator to find \( (1+i)^{4} \) and \( (1-i)^{4} \). After calculation, it holds that \( (1+i)^4 = 4 \) and \( (1-i)^4 = 4 \). thus, \((1+i)^{4}-(1-i)^{4} = 0 \).
03
Calculate \( (1+i)^{8} \) and \( (1-i)^{8} \)
Similar to Step 2, calculate \( (1+i)^{8} \) and \( (1-i)^{8} \). The result is \( (1+i)^8 = 16 \) and \( (1-i)^8 = 16\), making \( (1+i)^{8}-(1-i)^{8} = 0 \).
04
Calculate \( (1+i)^{12} \) and \( (1-i)^{12} \)
Find \( (1+i)^{12} \) and \( (1-i)^{12} \). It results in \( (1+i)^{12} = 64 \) and \( (1-i)^{12} = 64 \), hence, \( (1+i)^{12}-(1-i)^{12} = 0 \).
05
Predict \( (1+i)^{16} - (1-i)^{16} \)
By observing the results, the expression results in 0 for \(k = 4, 8, 12\). Therefore, it is logical to predict that \( (1+i)^{16} - (1-i)^{16} = 0 \) as well.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Imaginary Unit
The imaginary unit, often represented by the symbol \( i \), is a fundamental concept in the field of complex numbers. It is defined with the property that \( i^2 = -1 \). This means that \( i \) is not a real number, but when squared, it yields a real number: \(-1\). This property is the cornerstone of complex numbers and allows for the extension of real numbers into the complex plane.
Complex numbers are of the form \( a + bi \), where \( a \) and \( b \) are real numbers. Here, \( a \) is known as the real part, and \( bi \) is the imaginary part. The introduction of \( i \) into algebra makes it possible to solve equations that do not have solutions in the set of real numbers.
Complex numbers are of the form \( a + bi \), where \( a \) and \( b \) are real numbers. Here, \( a \) is known as the real part, and \( bi \) is the imaginary part. The introduction of \( i \) into algebra makes it possible to solve equations that do not have solutions in the set of real numbers.
- The standard notation for powers of \( i \) is cyclic. This means:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- This cycle repeats every four powers.
Binomial Theorem
The Binomial Theorem is a powerful tool for expanding expressions that are raised to a power. It provides a systematic way to contextualize the expansion of binomials, expressions of the form \( (a + b)^n \). The theorem is given by the formula:
\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]
Here, \( \binom{n}{k} \) is the binomial coefficient, representing the number of combinations of \( n \) items taken \( k \) at a time. The Binomial Theorem is especially helpful when dealing with complex numbers, as it allows for the expression of complex powers in expanded form.
\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]
Here, \( \binom{n}{k} \) is the binomial coefficient, representing the number of combinations of \( n \) items taken \( k \) at a time. The Binomial Theorem is especially helpful when dealing with complex numbers, as it allows for the expression of complex powers in expanded form.
- The application of the Binomial Theorem in the exercise involves evaluating expressions of the form \((1+i)^k\) and \((1-i)^k\).
- This theorem simplifies these complex expressions into simpler terms that can be individually calculated.
- By applying the theorem, one can break down complex expressions into manageable pieces, making calculations much more straightforward.
Power of Complex Numbers
The power of complex numbers involves raising a complex number, such as \((1+i)\), to a certain exponent, such as \(k\). Calculating the power of a complex number can be achieved through different methods, including the use of De Moivre's Theorem or direct multiplication.
For the exercise given, powers like \((1+i)^k\) and \((1-i)^k\) were calculated for different values of \(k\). A systematic approach is required:
For the exercise given, powers like \((1+i)^k\) and \((1-i)^k\) were calculated for different values of \(k\). A systematic approach is required:
- With the help of the Binomial Theorem, each term in the expansion can be evaluated.
- The expression is broken down into real and imaginary components, simplifying the computation.
- Observations on the recurring results for specific powers show patterns, as seen with results yielding zero for various values of \(k\).