Problem 43 - College Algebra with Corequisite Support

Question

For the following exercises, use a calculator to help answer the questions. Evaluate \((1-i)^{k}\) for \(k=2,6\) and \(10 .\) Predict the value if \(k=14\).

Step-by-Step Solution

Answer

\((1-i)^2 = -2i, (1-i)^6 = 8i, (1-i)^{10} = 32i, (1-i)^{14} = 128i\).

Step 1: Evaluate \((1-i)^2\)

Calculate \((1-i)^2\) using the formula for the square of a binomial. \((1-i)^2 = (1-i)(1-i) = 1^2 - 2 \cdot 1 \cdot i + (i)^2 = 1 - 2i + (-1) = 0 - 2i = -2i\).Thus, \((1-i)^2 = -2i\).

Step 2: Evaluate \((1-i)^6\)

First, recognize that \((1-i)^6 = ((1-i)^2)^3\). From Step 1, we know \((1-i)^2 = -2i\), so:\((-2i)^3 = (-2)^3 \cdot i^3 = -8 \cdot (-i) = 8i\).Thus, \((1-i)^6 = 8i\).

Step 3: Evaluate \((1-i)^{10}\)

First, use the property that \((1-i)^{10} = ((1-i)^2)^5\). From Step 1, we know \((1-i)^2 = -2i\), so:\((-2i)^5 = (-2)^5 \cdot i^5 = -32 \cdot (-i) = 32i\).Thus, \((1-i)^{10} = 32i\).

Step 4: Predict \((1-i)^{14}\)

Using the pattern from the powers that are multiples of 2, calculate \((1-i)^{14} = ((1-i)^2)^7\). Since \((1-i)^2 = -2i\), this becomes:\((-2i)^7 = (-2)^7 \cdot i^7 = -128 \cdot (-i) = 128i\).Thus, \((1-i)^{14} = 128i\).

Key Concepts

Binomial Theorem

The Binomial Theorem is a way to expand expressions raised to a power. In our problem, we are looking at expressions like \((1-i)^2\). To solve this, we can use the formula for the square of a binomial: \((a-b)^2 = a^2 - 2ab + b^2\).
Here, \(a = 1\) and \(b = i\). This theorem is super helpful because it simplifies calculations when expanding binomials raised to any power without multiplying out the expression repeatedly.For example: - When squaring \(1 - i\), applying the binomial formula gives us the steps seen in Step 1 of our solution: - Calculate each part following the formula. - Combine them to simplify to get \(-2i\).
This process helps break down the problem into smaller parts, making it straightforward.

Exponential Powers

Understanding Exponential Powers is crucial when working with expressions like \((1-i)^k\). An exponential power means raising a number or expression to a given power, e.g., squaring or cubing.In our example, we evaluate - \((1-i)^2\)- \((1-i)^6\)- \((1-i)^{10}\)For higher powers, notice how we simplify the problem by recognizing patterns or efficiently using what we've already calculated. For instance:- We used \((1-i)^2 = -2i\) in Step 1 and reused it to find - \((1-i)^6 = ((1-i)^2)^3\) which turns to \((-2i)^3\).The exponential approach saves lots of time by reducing repetitive calculations when the same small parts are multiplied multiple times.

Imaginary Unit

The Imaginary Unit, represented by \(i\), is a fundamental concept in complex numbers. It is defined by the property: - \(i^2 = -1\).
The role of the imaginary unit is essential in calculations involving complex numbers. Understanding how multiplying \(i\) affects numbers is key.In the solutions:- Recognize \(i^3 = -i\) because \(i^2\cdot i = (-1)\cdot i = -i\).- Similarly, \(i^5\) reduces to \(i\) because its cycle repeats every 4: - \(i, -1, -i, 1\), and back to \(i\) again. The cycle allows predicting results without recalculating each exponent case.

Patterns in Exponents

Identifying Patterns in Exponents is a big help with problems involving exponential powers. Patterns emerge when looking at the powers of complex numbers like \((1-i)^k\).Observe the examples:- For \(k=2,6,10,14\), we express the power as multiples of a simpler calculation, \((1-i)^2 = -2i\).
- By expressing higher powers in terms of \((1-i)^2\), we observe a repeating cycle. Each time, notice:- Every even power of \((1-i)\) is an imaginary number: - \(-2i, 8i, 32i\), etc.- This shows a consistent pattern across solved steps, as complex multiplication follows certain predictable outcomes.Such patterns are not only interesting but significantly simplify calculations when you see them early!