Problem 22 - Advanced Engineering Mathematics

Question

Write the given number in the form \(a+i b\). $$ (1+i)^{2}(1-i)^{3} $$

Step-by-Step Solution

Answer

The expression simplifies to \(4 - 4i\).

Step 1: Expand \((1+i)^2\)

First, expand the expression \((1+i)^2\) using the formula for squaring a binomial \((a+b)^2 = a^2 + 2ab + b^2\). Here, \(a = 1\) and \(b = i\). So, \((1+i)^2 = 1^2 + 2\times 1 \times i + i^2 = 1 + 2i + i^2\). Since \(i^2 = -1\), the expression becomes \(1 + 2i - 1 = 2i\).

Step 2: Expand \((1-i)^3\)

Next, expand \((1-i)^3\) using the binomial expansion formula \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\) with \(a = 1\) and \(b = i\). This gives us \(1^3 - 3\times 1^2 \times i + 3\times 1 \times i^2 - i^3\). Substitute \(i^2 = -1\) and \(i^3 = -i\): \(1 - 3i - 3 + i = -2 - 2i\).

Step 3: Multiply the results \(2i\) and \(-2 - 2i\)

Now, multiply the results from Steps 1 and 2: \(2i\times (-2 - 2i)\). Distribute \(2i\) across each term: \(2i \times -2 + 2i \times (-2i)\). This becomes \(-4i + 4\), since \((2i)\times (-2i) = 4\).

Step 4: Combine into the form \(a + ib\)

Finally, combine the real and imaginary parts to express the solution in the form \(a + ib\). From Step 3, we have \(-4i + 4\), which is written as \(4 + (-4i)\) or \(4 - 4i\).

Key Concepts

Binomial Expansion

Binomial expansion is a handy way to expand expressions that have powers, like \[(a+b)^n\]. It helps you break down these powers into simpler components. Here's how it works:

• When you square a binomial, such as \((a+b)^2\), you use the identity: \(a^2 + 2ab + b^2\).
• This tells us that when you expand \((1+i)^2\), you end up with \(1^2 + 2 imes 1 imes i + i^2\), leading to \(1 + 2i + i^2\).

Notice that the terms consist of simple powers and products that need to be simplified further. Since \(i^2 = -1\), the final expression for \((1+i)^2\) is \(2i\). Binomial expansion gets even more interesting with higher powers like \((1-i)^3\), where you utilize the formula: \(a^3 - 3a^2b + 3ab^2 - b^3\). Each of these terms represents a simple multiplication or squaring until the expression fully breaks down into simpler components that you can easily simplify.

Imaginary Unit

The imaginary unit, denoted as \(i\), is a fascinating concept in mathematics. It opens a world beyond the real numbers where you can calculate the square root of negative numbers.

• It is defined by the property that \(i^2 = -1\). This means that \(i\) is the square root of -1.
• Complex numbers are often expressed as \(a+ib\), where \(a\) and \(b\) are real numbers, and \(i\) signifies that imaginary component.

In computations such as the one we are examining, you frequently square \(i\), multiply it with other terms, or raise it to higher powers to simplify expressions. For example, \(i^3\) equals \(-i\), because \(i^3 = i^2 imes i = -1 imes i = -i\). Understanding how to manipulate \(i\) is crucial for working with complex numbers in mathematical problems.

Multiplication of Complex Numbers

Multiplying complex numbers involves distributing terms just as you do with algebraic expressions. The key difference is the need to account for \(i\) in your computations.

• Take the expressions \((2i)\) and \((-2 - 2i)\). You use distribution: \(2i imes -2 + 2i imes (-2i)\).
• First, \(2i imes -2 = -4i\). Multiplying a real number with an imaginary, you keep the \(i\) element.
• Second, \(2i imes (-2i) = 4\), because \(i^2 = -1\). The negative sign converts \(-4i^2\) to \(+4\).

Bring it all together, and the resulting complex number is \(4 - 4i\). Multiplying complex numbers becomes straightforward once all terms are considered, simplified, and combined correctly. Knowing how to manage the presence of \(i\), especially during multiplication, ensures accuracy and ease in maneuvering through calculations.