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The distributive property is a fundamental concept in algebra that allows us to multiply a sum by distributing the multiplication over each term within the parentheses. For example, when we see the expression \((a + b)(c + d)\), this multiplication distributes as \((a \cdot c + a \cdot d + b \cdot c + b \cdot d)\).
With complex numbers, the process is similar, but you need to carefully track the imaginary unit \(i\) while performing these operations.
In the exercise \((1 - i)(1 + i)\), applying the distributive property means multiplying each term in \((1 - i)\) by each term in \((1 + i)\).
Therefore, the result is:

• \(1 \cdot 1\)
• \(1 \cdot i\)
• \(-i \cdot 1\)
• \(-i \cdot i\).

This breakdown ensures all components of the complex numbers are multiplied correctly.

Simplifying expressions in complex numbers is essential after distributing terms. The main goal here is to reduce the expression to a simpler form.
From \( (1 - i)(1 + i) \), we obtain: \(1 + i - i - i^2\).
The terms \(1 \cdot 1,\) \(1 \cdot i,\) \(-i \cdot 1,\) and \(-i \cdot i\) were computed during distribution and the result was gathered as \(1 + i - i - i^2\).
The imaginary parts \(i\) and \(-i\) cancel each other since \(i - i = 0\).
So simplifying further leads us to handle the \(-i^2\) term next.

The imaginary unit, denoted as \(i,\) is a special constant with the property that \(i^2 = -1\).
This property is crucial in simplifying complex number expressions.
In the expression \(1 + i - i - i^2,\) we know \(-i^2\) actually translates to \(+1\) because \(i^2 = -1\) implies \(-i^2 = -(-1) = 1\).
Remembering this crucial property allows us to modify \(-i^2\) into a real number addition, changing the expression to \(1 + 1\).
Understanding how \(i\) works makes handling complex operations simpler.

Algebraic operations on complex numbers often involve combining like terms to reach a simplified result.
Like terms in the context of complex numbers are those involving the imaginary unit \(i\) and those that do not.
When the expression \(1 + i - i + 1\) is simplified and results in \(1 + 1 = 2\), we're left purely with a real number.
The imaginary components \(i\) and \(-i\) cancel each other out, resulting in zero.
After canceling and combining like terms, we are left with the simplified form \(2 + 0i\), which is the final form \(a + bi\) of the expression.
The beauty of algebraic operations lies in transforming a seemingly complex math problem into a straightforward answer.