1. **State the problem:** Separate the expression $$\frac{(-2+3i)^2}{1+i}$$ into its real and imaginary parts. 2. **Recall formulas 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. - The conjugate of $1+i$ is $1 - i$. 3. **Calculate the numerator:** $$(-2+3i)^2 = (-2)^2 + 2 \times (-2) \times 3i + (3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i$$ 4. **Multiply numerator and denominator by conjugate of denominator:** $$\frac{-5 - 12i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(-5 - 12i)(1 - i)}{(1 + i)(1 - i)}$$ 5. **Calculate denominator:** $$(1 + i)(1 - i) = 1 - i + i - i^2 = 1 - (-1) = 1 + 1 = 2$$ 6. **Calculate numerator:** $$(-5)(1) + (-5)(-i) + (-12i)(1) + (-12i)(-i) = -5 + 5i - 12i + 12i^2 = -5 - 7i + 12(-1) = -5 - 7i - 12 = -17 - 7i$$ 7. **Divide numerator by denominator:** $$\frac{-17 - 7i}{2} = -\frac{17}{2} - \frac{7}{2}i$$ 8. **Result:** - Real part: $$-\frac{17}{2}$$ - Imaginary part: $$-\frac{7}{2}$$ **Final answer:** $$\boxed{-\frac{17}{2} - \frac{7}{2}i}$$