Problem 37 - Contemporary Precalculus

Question

Expand and (where possible) simplify the expression. $$(1+i)^{6}, \text { where } i^{2}=-1$$

Step-by-Step Solution

Answer

Answer: The simplified expression is 32.

Step 1: Use the binomial theorem to expand the expression

The binomial theorem states that for any natural number \(n\) and complex numbers \(a\) and \(b\), $$(a+b)^n = \sum_{k=0}^n {n\choose k} a^{n-k}b^{k} = {n\choose0}a^n + {n\choose1}a^{n-1}b + \cdots + {n\choose n} b^n$$ In our case, we have \(a = 1\), \(b = i\), and \(n = 6\). Applying the binomial theorem, we get: $$(1+i)^{6} = \sum_{k=0}^6 {6\choose k} 1^{6-k}i^{k}$$

Step 2: Substitute the binomial coefficients and simplify the powers of i

Now, we need to substitute the binomial coefficients for each term and simplify the powers of \(i\). We can use the property \(i^2 = -1\) to simplify higher powers of \(i\). Here's the expanded form and simplification of \((1+i)^{6}\): $$(1+i)^{6} = {6\choose0} + {6\choose1}i + {6\choose2}i^2 + {6\choose3}i^3 + {6\choose4}i^4 + {6\choose5}i^5 + {6\choose6}i^6$$ $$ = 1 + 6i - 15i^2 - 20i^3 + 15i^4 + 6i^5 - i^6$$ Now, we can use the fact that \(i^2 = -1\) and that \(i^3 = i^2 \cdot i = -i\) and \(i^4 = i^2 \cdot i^2 = 1\) to simplify further: $$ = 1 + 6i - 15(-1) - 20(-i) + 15(1) + 6i^3 - i^4$$ $$ = 1 + 6i + 15 - 20i + 15 - 6i + 1$$ $$ = (1 + 15 + 15 + 1) + (-20i + 6i - 6i)$$ $$ = 32$$

Step 3: Write the final answer

After expanding and simplifying the expression \((1+i)^{6}\) using the binomial theorem and the given property of \(i^2 = -1\), we found that: $$(1+i)^{6} = 32$$

Key Concepts

Complex Numbers

Complex numbers are numbers that have a real and an imaginary part. The standard form of a complex number is written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part, and \(i\) is the imaginary unit. The imaginary unit \(i\) is defined by the property that \(i^2 = -1\). This means that complex numbers can be ideal for calculations involving roots of negative numbers, something not possible with only real numbers.

• Real part (\(a\)): Represents the real number segment of a complex number.
• Imaginary part (\(bi\)): Represents the imaginary portion, where \(i\) is the square root of -1.

Operations with complex numbers such as addition, subtraction, multiplication, and division follow certain rules stemming from the properties of \(i\). For example, when multiplying complex numbers, like \((a + bi)(c + di)\), we use distributive property while keeping in mind that \(i^2 = -1\).

This foundational understanding of complex numbers becomes very useful in more advanced topics such as the expansion of powers as seen in exercises like \((1+i)^6\).

Powers of Complex Numbers

Calculating powers of complex numbers involves repeated multiplication, and because of the nature of \(i\), it follows a cyclical pattern. The powers of \(i\) cycle every four powers:

• \(i^0 = 1\)
• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\) (and the cycle repeats)

This repeatedly cyclical property makes simplifying expressions involving powers of \(i\) straightforward after the cycle is recognized. For instance, in expanding and simplifying \((1+i)^6\), understanding that \(i^4 = 1\) makes it easier to deal with terms like \(i^5\) or \(i^6\) efficiently.

Using the binomial theorem — which helps in expanding expressions raised to the power \(n\) — combined with the cyclical pattern of \(i\)'s powers, allows for the effective simplification of complex expressions, making problems like finding \((1+i)^6\) manageable and straightforward.

Algebraic Expansion

Algebraic expansion involves expanding expressions such as \((a+b)^n\) into a sum of terms using techniques like the binomial theorem. This theorem is crucial when working with powers of binomials, where you expand it into a series of terms according to coefficients determined by combinations (also known as binomial coefficients).

In our example \((1+i)^6\), the binomial theorem is used:\[\sum_{k=0}^6 {6\choose k} 1^{6-k}i^{k}\]This means you expand the binomial into 7 terms (0 to 6), each involving coefficients from Pascal's Triangle, and powers of 1 and \(i\). When performing expansions like this:

• Calculate the binomial coefficients, which are combinations \({n\choose k}\).
• Apply powers accordingly (like \(1^{6-k}\) which simplifies to 1).
• Consider the cycles of \(i\)'s powers to simplify the imaginary terms.

Thus, from an algebraic expansion viewpoint, knowing both the theorem and the properties of components like complex numbers makes the simplification process efficient and understandeable.