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The difference of squares is a mathematical identity used to simplify expressions that are in the form \((a-b)(a+b)\). This specific form leads to a simplified result using the formula \((a-b)(a+b) = a^2 - b^2\). The expression resembles a pattern where the product involves the same two terms but with opposite signs. By applying the difference of squares, calculations become more manageable.

• Formula: The general formula is \((a-b)(a+b) = a^2 - b^2\).
• Usage: It allows the multiplication of conjugate pairs, like \((\sqrt{5} - 5i)(\sqrt{5} + 5i)\), to be done quickly by calculating just the squares of each component.

In our exercise, \(a\) is \(\sqrt{5}\) and \(b\) is \(5i\), so the expression can be instantly simplified by applying \(a^2 - b^2\). Understanding these shortcuts allows one to solve multiplications involving complex numbers without expanding the entire expression, saving time and effort.

The imaginary unit, denoted by \(i\), is a fundamental concept in complex numbers. The distinguishing feature of \(i\) is its property such that \(i^2 = -1\). This allows mathematicians and students to handle square roots of negative numbers, making it possible to perform calculations that real numbers alone cannot.

• Definition: \(i\) is defined such that \(i^2 = -1\).
• Usage: It is essential in formulating complex numbers, written as \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.
• Implications: Understanding \(i\) enables calculation of powers of \(i\) using its cyclical properties: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\), which then repeats.

In the context of the exercise, identifying that \((5i)^2 = 25i^2 = -25\) is based on the property \(i^2 = -1\). Thus, understanding the imaginary unit simplifies calculations involving complex numbers.

Multiplying complex numbers can seem overwhelming initially but is straightforward with practice. A complex number is generally expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers. The multiplication utilizes the distributive property, combined with the properties of the imaginary unit, \(i\).In our exercise, we multiplied \((\sqrt{5} - 5i)(\sqrt{5} + 5i)\). Let's break down how complex multiplication works:

• Distributive Property: Apply \'foil\' – First, Outside, Inside, Last, which yields results for each possible product between terms.
• Simplification: Combine like terms and use \(i^2 = -1\) to simplify the expression.

However, since our expression is a difference of squares, we used \(a^2 - b^2\) directly, simplifying it to 30 immediately. Understanding this powerful shortcut not only aids in solving the problem more efficiently but also in handling any multiplication of complex numbers involving conjugate pairs effectively.