If $\alpha$ and $\beta$ are roots of $z^2+4z+8=0$, then $\dfrac{\alpha+\beta+4i}{\alpha\beta+8i}$ in simplest form is:

1. For $z^2+bz+c=0$ the sum of roots is $-b$ and the product is $c$. 2. So $\alpha+\beta=-4$ and $\alpha\beta=8$, with no need to solve the quadratic. 3. Numerator: $-4+4i=4(i-1)$. 4. Denominator: $8+8i=8(1+i)$. 5. The ratio is $\dfrac{4(i-1)}{8(1+i)}=\dfrac{i-1}{2(1+i)}$; multiplying top and bottom by $(1-i)$ gives $\dfrac{2i}{4}=\dfrac{i}{2}$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 4: Complex Numbers_