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Expanding complex numbers is much like expanding binomials in algebra, but with one crucial difference: the presence of the imaginary unit, denoted as 'i'. Complex numbers are in the form of 'a + bi', where 'a' is the real part and 'b' is the imaginary part. When expanding, like in the exercise \( (5 - 4i)^2 \), we treat 'i' just as we would any variable.

However, a key point of difference is how we handle powers of 'i'. For example: \((i^2 = -1)\), \((i^3 = -i)\), and \((i^4 = 1)\). During expansion, we apply the distributive property to complex numbers in the same way we do with real numbers. Here we use the FOIL method to expand \( (5 - 4i)(5 - 4i) \) which results in \(25 - 20i - 20i + 16i^2\). By keeping track of every 'i' during expansion, we ensure an accurate simplification when combining like terms thereafter.

The distributive property is an algebraic rule for multi-term expressions which dictates how to multiply a single term by a sum or difference. In the context of complex numbers, the distributive property is applied by multiplying each part of one complex number by each part of the other. In the expansion of \( (5 - 4i)^2 \), it translated to finding the product of each possible pair of terms: one from each binomial.

Here's a step-by-step breakdown using the distributive property:

• First pair: \(5 \times 5 = 25\)
• Second pair: \(5 \times -4i = -20i\)
• Third pair: \(5 \times -4i = -20i\) again
• Fourth pair: \( -4i \times -4i = 16i^2\)

After applying the distributive property, we combine like terms to simplify our expression before substituting \(i^2\) with \( -1\).

Arithmetic involving the imaginary unit 'i' follows its own set of rules, reflective of \(i^2 = -1\). It's pivotal to handle powers of 'i' correctly in the simplification of expanded complex numbers. As seen in the exercise, after the initial expansion, we get a term \(16i^2\), which we can simplify using the fact that \(i^2 = -1\).

This substitution is the next step post expansion and greatly impacts the simplification process. An error in this step can lead to an incorrect standard form. After the substitution, \(16i^2\) gets simplified to -16, and we can combine this with the other real numbers in the expression. The accuracy of the imaginary arithmetic is, therefore, crucial in obtaining the standard form, represented as \(a + bi\), of the complex number after expansion and combination of like terms.