1. **Problem:** Write each expression in the standard form $a + bi$. **a)** $(-3 - 2i)^2 + \frac{1 + i}{8 - 5i}$ 2. **Formula and rules:** - To square a complex number: $(a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2abi$ because $i^2 = -1$. - To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator. 3. **Step-by-step solution:** - Square $(-3 - 2i)$: $$(-3 - 2i)^2 = (-3)^2 + 2(-3)(-2i) + (-2i)^2 = 9 + 12i + 4i^2 = 9 + 12i - 4 = 5 + 12i$$ - Simplify $\frac{1 + i}{8 - 5i}$ by multiplying numerator and denominator by conjugate $8 + 5i$: $$\frac{1 + i}{8 - 5i} \times \frac{8 + 5i}{8 + 5i} = \frac{(1 + i)(8 + 5i)}{8^2 + 5^2} = \frac{8 + 5i + 8i + 5i^2}{64 + 25} = \frac{8 + 13i - 5}{89} = \frac{3 + 13i}{89} = \frac{3}{89} + \frac{13}{89}i$$ - Add the two results: $$5 + 12i + \frac{3}{89} + \frac{13}{89}i = \left(5 + \frac{3}{89}\right) + \left(12 + \frac{13}{89}\right)i = \frac{445}{89} + \frac{1081}{89}i$$ **Final answer a):** $\frac{445}{89} + \frac{1081}{89}i$ --- **b)** $6i^{31} - 2i(7 + 11i)$ - Recall powers of $i$: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and cycle repeats every 4. - Find $i^{31}$: $31 \mod 4 = 3$, so $i^{31} = i^3 = -i$. - Calculate: $$6i^{31} = 6(-i) = -6i$$ $$-2i(7 + 11i) = -2i \times 7 - 2i \times 11i = -14i - 22i^2 = -14i - 22(-1) = -14i + 22$$ - Sum: $$-6i + (-14i + 22) = 22 - 20i$$ **Final answer b):** $22 - 20i$ --- **c)** $(2 - 3i)^3$ - Use binomial expansion or cube formula: $$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$ - Let $a=2$, $b=3i$: $$2^3 - 3 \times 2^2 \times 3i + 3 \times 2 \times (3i)^2 - (3i)^3 = 8 - 36i + 3 \times 2 \times 9 i^2 - 27 i^3$$ - Calculate powers: $$i^2 = -1, \quad i^3 = i^2 \times i = -1 \times i = -i$$ - Substitute: $$8 - 36i + 54(-1) - 27(-i) = 8 - 36i - 54 + 27i = (8 - 54) + (-36i + 27i) = -46 - 9i$$ **Final answer c):** $-46 - 9i$