1. **State the problem:** Evaluate the expression $ (1 - i)^6 $ where $i$ is the imaginary unit with $i^2 = -1$. 2. **Recall the formula and rules:** To simplify powers of complex numbers, it is often easier to convert to polar form or use binomial expansion. Here, we use the binomial theorem: $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ where $a = 1$, $b = -i$, and $n = 6$. 3. **Apply the binomial theorem:** $$ (1 - i)^6 = \sum_{k=0}^6 \binom{6}{k} 1^{6-k} (-i)^k = \sum_{k=0}^6 \binom{6}{k} (-1)^k i^k $$ 4. **Calculate each term:** - $\binom{6}{0} (-1)^0 i^0 = 1$ - $\binom{6}{1} (-1)^1 i^1 = 6 \times (-1) \times i = -6i$ - $\binom{6}{2} (-1)^2 i^2 = 15 \times 1 \times (-1) = -15$ - $\binom{6}{3} (-1)^3 i^3 = 20 \times (-1) \times (-i) = 20i$ - $\binom{6}{4} (-1)^4 i^4 = 15 \times 1 \times 1 = 15$ - $\binom{6}{5} (-1)^5 i^5 = 6 \times (-1) \times i = -6i$ - $\binom{6}{6} (-1)^6 i^6 = 1 \times 1 \times (-1) = -1$ 5. **Sum real and imaginary parts:** Real parts: $1 - 15 + 15 - 1 = 0$ Imaginary parts: $-6i + 20i - 6i = 8i$ 6. **Final answer:** $$ (1 - i)^6 = 8i $$ This means the value is purely imaginary with magnitude 8 and no real part.