Problem 44 - College Algebra with Corequisite Support

Question

For the following exercises, 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 \(k=16\).

Step-by-Step Solution

Answer

The expression evaluates to 0 for all given k and predicted k values.

Step 1: Calculate \((1+i)^4\)

To calculate \((1+i)^4\), we apply the binomial theorem which gives us: \[(1+i)^4 = \sum_{n=0}^{4} \binom{4}{n} (1)^{4-n} (i)^n\]This expands to \((1+i)^4 = 1^4 + 4 \times 1^3 \times i + 6 \times 1^2 \times (i^2) + 4 \times 1 \times (i^3) + i^4\), which simplifies to:\((1+i)^4 = 1 + 4i - 6 - 4i + 1 = -4.\)

Step 2: Calculate \((1-i)^4\)

Similarly, expand \((1-i)^4\) using the binomial theorem:\[(1-i)^4 = \sum_{n=0}^{4} \binom{4}{n} (1)^{4-n} (-i)^n\]Expanding gives:\((1-i)^4 = 1 - 4i - 6 + 4i + 1 = -4.\)

Step 3: Compute \((1+i)^4 - (1-i)^4\)

Now subtract the results from Steps 1 and 2:\((1+i)^4 - (1-i)^4 = -4 - (-4) = 0.\)

Step 4: Calculate \((1+i)^8\)

Now apply the same method for \((1+i)^8\):Realize that \((1+i)^8 = [(1+i)^4]^2 = (-4)^2 = 16.\)

Step 5: Calculate \((1-i)^8\)

Similarly, calculate for \((1-i)^8\):\((1-i)^8 = [(1-i)^4]^2 = (-4)^2 = 16.\)

Step 6: Compute \((1+i)^8 - (1-i)^8\)

Subtract the results from Steps 4 and 5:\((1+i)^8 - (1-i)^8 = 16 - 16 = 0.\)

Step 7: Calculate \((1+i)^{12}\)

Using the same approach,\((1+i)^{12} = [(1+i)^4]^3 = (-4)^3 = -64.\)

Step 8: Calculate \((1-i)^{12}\)

Similarly for \((1-i)^{12}\):\((1-i)^{12} = [(1-i)^4]^3 = (-4)^3 = -64.\)

Step 9: Compute \((1+i)^{12} - (1-i)^{12}\)

Subtract the results from Steps 7 and 8:\((1+i)^{12} - (1-i)^{12} = -64 - (-64) = 0.\)

Step 10: Predict for \((1+i)^{16} - (1-i)^{16}\)

Based on the pattern observed (that each calculation results in 0), we predict:\((1+i)^{16} - (1-i)^{16} = 0.\) This is due to the symmetry in powers of complex conjugates combining to yield real terms.

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool used to expand expressions raised to a certain power. For an expression of the form \((a+b)^n\), the theorem states that it can be expanded as:

• \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

This means we are summing up all the terms, where each term is a product of binomial coefficients \(\binom{n}{k}\), power of \(a\), and power of \(b\).
For complex numbers, this theorem is helpful because it allows us to systematically expand expressions like \((1+i)^k\) or \((1-i)^k\), both of which involve imaginary unit \(i\), where \(i^2 = -1\).
Example: To compute \((1+i)^4\) using the binomial theorem, we apply \[ \sum_{n=0}^{4} \binom{4}{n} (1)^{4-n} (i)^n \]which simplifies using the properties of \(i\).
Thus, understanding the Binomial Theorem is crucial for such calculations as it breaks down complex expressions into manageable pieces.

Complex Conjugates

Complex conjugates are pairs of complex numbers that have the same real part but opposite imaginary parts. If a complex number is represented as \(a+bi\), its conjugate is \(a-bi\).
The beauty of complex conjugates lies in their properties:

• Multiplying a complex number by its conjugate results in a real number: \((a+bi)(a-bi) = a^2+b^2\).
• The sum and the product of complex conjugates are real numbers. This is evident in the original exercise, showing that expressions like \((1+i)^k\) and \((1-i)^k\) combine (or subtract) to yield zero for certain powers.

Understanding this concept helps predict behaviors of complex expressions, just like \((1+i)^{16} - (1-i)^{16} = 0\), and highlights the symmetric nature of complex numbers.

Powers of Complex Numbers

Taking powers of complex numbers involves repeatedly multiplying the base by itself, and often using identities like \(i^2 = -1\).
For example, calculating \((1+i)^k\) can involve the Binomial Theorem initially, but once known, earlier powers can simplify future calculations:

• \((1+i)^8 = ((1+i)^4)^2 = (-4)^2 = 16\)
• \((1+i)^{12} = ((1+i)^4)^3 = (-4)^3 = -64\)

A pattern emerges with higher powers, helping make predictions for powers like \((1+i)^{16}\) without re-expanding from scratch. Complex numbers thus not only enrich mathematical explorations within geometry but also through algebraic symmetries, enhancing our problem-solving toolkit.