Problem 43 - College Algebra

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)^{14} = 128i.

Step 1: Evaluate for k=2

To evaluate \((1-i)^2\), we start by using the binomial expansion formula: \((1-i)^2 = (1)^2 + 2(1)(-i) + (-i)^2 = 1 - 2i + i^2\). Since \(i^2 = -1\), this simplifies to \(1 - 2i - 1 = -2i\). Thus, \((1-i)^2 = -2i\).

Step 2: Evaluate for k=6

Write \((1-i)^6 = ((1-i)^2)^3\). We found \((1-i)^2 = -2i\) in step 1, so substitute: \((1-i)^6 = (-2i)^3\). Calculate \((-2i)^3 = -8i^3\). Since \(i^3 = -i\), substitute to get \(-8i^3 = -8(-i) = 8i\). So, \((1-i)^6 = 8i\).

Step 3: Evaluate for k=10

First express \((1-i)^{10} = ((1-i)^2)^5\). From step 1, \((1-i)^2 = -2i\). So, \((1-i)^{10} = (-2i)^5\). Calculate \((-2i)^5 = -32i^5\). Given \(i^5 = i\), substitute: \(-32i^5 = -32i\). Therefore, \((1-i)^{10} = -32i\).

Step 4: Predict the value for k=14

Due to the pattern observed, calculate \((1-i)^{14} = ((1-i)^2)^7\). With \((1-i)^2 = -2i\), substitute to find \((1-i)^{14} = (-2i)^7\). Computing \((-2i)^7 = -128i^7\) and knowing \(i^7 = -i\), replace to get \(-128i^7 = -128(-i) = 128i\). Therefore, predict \((1-i)^{14} = 128i\).

Key Concepts

Binomial Expansion

Binomial expansion is a critical concept used when raising expressions with two terms, like \((1-i)\), to a power. It involves expanding the expression by applying the binomial theorem, which uses the formula \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). For complex numbers, like \((1-i)^2\), the formula helps us to expand the expression systematically.

In the exercise, we start with \((1-i)^2\). This means we treat it as \((a-b)^2\), where \(a=1\) and \(b=i\). Using the formula:

• First term: \(1^2 = 1\),
• Middle term: \(2 \cdot 1 \cdot (-i) = -2i\),
• Last term: \((-i)^2 = i^2 = -1\).

This expansion gives: \(1 - 2i - 1\), which simplifies to \(-2i\). Hence, using binomial expansion allows us to break down the problem into manageable steps.

Powers of i

Understanding the powers of \(i\) (the imaginary unit) is essential in dealing with complex numbers. The imaginary unit \(i\) represents \(\sqrt{-1}\), and as \(i\) cycles through powers, it has a predictable pattern:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)
• The cycle repeats: \(i^5 = i\), and so on.

This behavior can simplify calculations significantly, as seen in the exercise when evaluating higher powers. For example, \(i^3 = -i\) helps to compute \((-2i)^3\), leading quickly to \(8i\). Knowing these cycles allows students to work through complex exponentiation more effectively.

Complex Exponentiation

Complex exponentiation refers to raising a complex number to a power. By using known properties of arithmetic, we can simplify expressions like \((1-i)^k\) into easier forms. In the exercise, deconstructing a power like \((1-i)^{10}\) helps manage complexity by first recognizing the expression as \(((1-i)^2)^5\) then using the result from the simpler \((1-i)^2 = -2i\).

This strategy aids in breaking down higher powers into repeated operations of lower powers already calculated. Thus, \((-2i)^5\) rests on the previous solution \((-2i)\), quickly leading to \(-32i\) by recognizing previously calculated outcomes and applying known \(i\) power patterns.

Pattern Recognition in Powers

Pattern recognition in powers is an invaluable skill in addressing problems involving exponentiation, especially with imaginary numbers. This technique is about spotting regularities and exploiting cycles of repetitive behavior, as seen with the powers of \(i\). In the given problem, recognizing the effect of repeated squaring helps to predict higher order power results like \((1-i)^{14}\).

We notice the outcomes for powers \(k=2, 6, 10\), and then predict \(k=14\) using observation: every further pair of power increments cycles through similar complex outcomes due to the characteristics of \((1-i)\). Thus, understanding and harnessing these patterns allow predictions about the powers of complex expressions without necessarily calculating each power step directly.