1. **State the problems:** We need to simplify the following complex number fractions: ii) $$\frac{(-2 + 3i)^2}{1 + i}$$ iii) $$\frac{i}{1 + i}$$ 2. **Recall important 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 to remove the imaginary part from the denominator. - The conjugate of $a + bi$ is $a - bi$. 3. **Simplify problem ii:** - First, square the numerator: $$(-2 + 3i)^2 = (-2)^2 + 2 \times (-2) \times 3i + (3i)^2 = 4 - 12i + 9i^2$$ - Since $i^2 = -1$, this becomes: $$4 - 12i + 9(-1) = 4 - 12i - 9 = -5 - 12i$$ - Now, divide by $1 + i$: $$\frac{-5 - 12i}{1 + i}$$ - Multiply numerator and denominator by the conjugate of the denominator $1 - i$: $$\frac{(-5 - 12i)(1 - i)}{(1 + i)(1 - i)}$$ - Calculate numerator: $$(-5)(1) + (-5)(-i) + (-12i)(1) + (-12i)(-i) = -5 + 5i - 12i + 12i^2$$ - Simplify numerator: $$-5 - 7i + 12(-1) = -5 - 7i - 12 = -17 - 7i$$ - Calculate denominator: $$(1)^2 - (i)^2 = 1 - (-1) = 2$$ - So the fraction is: $$\frac{-17 - 7i}{2} = -\frac{17}{2} - \frac{7}{2}i$$ 4. **Simplify problem iii:** - Start with: $$\frac{i}{1 + i}$$ - Multiply numerator and denominator by the conjugate of the denominator $1 - i$: $$\frac{i(1 - i)}{(1 + i)(1 - i)}$$ - Calculate numerator: $$i \times 1 - i \times i = i - i^2 = i - (-1) = i + 1$$ - Calculate denominator: $$1 - (-1) = 2$$ - So the fraction is: $$\frac{1 + i}{2} = \frac{1}{2} + \frac{1}{2}i$$ **Final answers:** ii) $$-\frac{17}{2} - \frac{7}{2}i$$ iii) $$\frac{1}{2} + \frac{1}{2}i$$