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To understand how to compute expressions like \[(1-i)^k\]it's essential to dive into the binomial expansion. It's a method used to expand expressions that are raised to any power. In simpler terms, it's a formula that helps quickly multiply out expressions of the form \((a-b)^n\).
The general expression for binomial expansion is given by:\[(a+b)^n = \sum_{{k=0}}^{n} \binom{n}{k} a^{n-k} b^k\]Here's what each component means:

• \(n\) is the power you're raising the binomial to.
• \(\binom{n}{k}\) is called a binomial coefficient, and it represents the number of ways to choose \(k\) elements from \(n\), calculated as \(\frac{n!}{k!(n-k)!}\).
• \(a\) and \(b\) are the two terms in the binomial, and \(k\) is the specific term in the expansion.

To solve \((1-i)^2\), we apply this expansion method, simplifying the terms where needed. In this exercise, since the binomial is \(1-i\), we fill in the formula with \(a = 1\) and \(b = -i\). This results in:\[(1-i)^2 = 1 - 2i + i^2\] where \(i^2 = -1\). This step-by-step expansion helps break down complex calculations into more manageable parts.

Understanding the powers of the imaginary unit, \(i\), can simplify computations extensively. The imaginary unit \(i\) is defined such that:

• \(i^1 = i\) – This is just \(i\) itself.
• \(i^2 = -1\) – Since \(i\) is defined as the square root of \(-1\).
• \(i^3 = -i\) – This is \(i^2 imes i = -1 imes i\).
• \(i^4 = 1\) – It equals \(i^2 imes i^2 = (-1) imes (-1)\).

Beyond that, the powers follow a repeating cycle every four exponents. Therefore, for higher powers like \(i^5\), you can simplify by first dividing the exponent by 4 and focus on the remainder. Here, \(i^5 = i^{4+1} = 1 imes i = i\). Using these identities, you can effortlessly calculate things like \((1-i)^6\) by first evaluating the power of \(i\) involved.

When dealing with complex numbers and raising them to higher powers, identifying patterns can reveal shortcuts and reduce computational effort. In the original problem, a pattern emerges when evaluating powers of \((1-i)\):
- \((1-i)^2 = -2i\)- \((1-i)^6 = 8i\)- \((1-i)^{10} = -32i\)Noticeably, two main patterns are observed:

• The coefficient of \(i\) in these results doubles each time we go to the next power of \((1-i)\) by 4.
• The sign alternates between positive and negative.

Thus, understanding these patterns affirms the prediction strategy used in the original exercise. For example, moving to \((1-i)^{14}\), using the alternating pattern, the result becomes positive again, and multiplying the coefficient by 4 yields \(128i\). Comprehending these patterns not only simplifies calculations but also deeply enriches understanding of the behavior of powers in complex number arithmetic.