Problem 12 - Advanced Engineering Mathematics

Question

Write the given number in the form \(a+i b\). $$ (1-i)^{3} $$

Step-by-Step Solution

Answer

The expression \((1-i)^3\) simplifies to \(-2-2i\).

Step 1: Expand the Expression

To find the cube of a binomial, we use the binomial theorem formula: \[ (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 \]In this context, let \( x = 1 \) and \( y = i \), so the expression becomes:\[ (1 - i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 \]

Step 2: Compute the Terms

Let's compute each term individually:1. \( 1^3 = 1 \)2. \( -3(1)^2(i) = -3i \)3. \( 3(1)(i)^2 = 3(i^2) \) where \( i^2 = -1 \), so this becomes \( 3(-1) = -3 \)4. \( -i^3 = -(-i) \) because \( i^3 = -i \). Thus, this becomes \( i \).

Step 3: Simplify the Expression

Combine all the terms computed in Step 2:\[ 1 - 3i - 3 + i \]Combine like terms:- The real parts: \( 1 - 3 = -2 \)- The imaginary parts: \( -3i + i = -2i \)Thus, the expression simplifies to \( -2 - 2i \).

Step 4: Express in Standard Form

Ensure the solution is in the form \( a + ib \), where \( a \) and \( b \) are real numbers. From the simplification, the expression is:\[ -2 - 2i \]This is already in the form \( a + ib \), where \( a = -2 \) and \( b = -2 \).

Key Concepts

Binomial Theorem

To understand how to expand expressions like (1-i)^3, we use the Binomial Theorem. This theorem is a powerful tool in algebra that helps to expand expressions raised to a power, especially binomials, which are expressions with two terms. It states that for any positive integer n, (x + y)^n can be expanded using the sum of terms \( \binom{n}{k} x^{n-k} y^k \) for k ranging from 0 to n.

• The coefficients \( \binom{n}{k} \) are calculated using combinations, often referred to as binomial coefficients. These give us the number of ways to select k items from a total of n.
• The pattern follows Pascal's Triangle, where coefficients are arranged in a triangle with each number being the sum of the two directly above it.

By substituting the appropriate values for x, y, and n, you can compute each term in the expansion and combine them together to get the final expanded result.

Imaginary Unit

The imaginary unit, denoted by i, is a fundamental element in complex number theory. It is defined such that \( i^2 = -1 \). This property allows us to extend the real number system to include numbers that have a real and imaginary part. When working with i, it’s crucial to remember its cyclical nature:

• \( i^0 = 1 \) – The zeroth power of i is 1, just like any number.
• \( i^1 = i \) – The first power is the imaginary unit itself.
• \( i^2 = -1 \) – By definition of the imaginary unit.
• \( i^3 = -i \) – Derived from multiplying \( i^2 \) by i.
• \( i^4 = 1 \) – Cycles back to one, repeating every four powers.

Understanding these powers helps simplify expressions involving complex numbers, as seen in the step-by-step reduction of \((1-i)^3\). Using the cyclical nature of i greatly simplifies calculations, making it an integral part of algebra involving complex numbers.

Polynomial Expansion

When you expand a polynomial, like a binomial raised to a power, you break it down into simpler terms. Each term of the polynomial is a simpler expression that separately contributes to the whole.
For the expression \((1-i)^3\), using the binomial expansion method involves realizing:

• A polynomial is made up of several terms. For example, \((x-y)^3\) expands to four terms.
• Each term is made up of a combination of the original terms raised to varying powers, reflecting their contribution to the expansion based on the binomial theorem coefficients.
• The expansions incorporate changes due to the imaginary unit, reflecting how imaginary powers interact with real terms and each other to adjust their signs and magnitudes.

This process reflects how polynomial expansion helps simplify and solve complex expressions, turning them into manageable algebraic pieces that can be easily combined and reduced, as seen when we simplified the expression to \-2 - 2i.\ Using polynomial expansion in conjunction with complex numbers allows us to understand and manipulate expressions perfectly suited for imaginary number calculations.