(1-i)³

To calculate \((1 - i)^3\), we can use the binomial theorem or expand it directly. 1. First, we can expand \((1 - i)^3\) using the binomial expansion formula: \[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \] In our case, \(a = 1\), \(b = i\), and \(n = 3\): \[ (1 - i)^3 = \sum_{k=0}^{3} \binom{3}{k} (1)^{3-k} (-i)^k \] Calculating the terms in the expansion: - For \(k = 0\): \(\binom{3}{0}(1)^{3}(−i)^{0} = 1\) - For \(k = 1\): \(\binom{3}{1}(1)^{2}(−i)^{1} = 3 \cdot (-i) = -3i\) - For \(k = 2\): \(\binom{3}{2}(1)^{1}(−i)^{2} = 3 \cdot (-1) = -3\) - For \(k = 3\): \(\binom{3}{3}(1)^{0}(−i)^{3} = 1 \cdot (-i^3) = 1 \cdot i = i\) Now combine these results: \[ (1 - i)^3 = 1 - 3i - 3 + i \] Combine the real and imaginary parts: \[ = (1 - 3) + (-3i + i) = -2 - 2i \] Therefore, the answer is: \[ \boxed{-2 - 2i} \]