If $\alpha+i\beta$ and $\gamma+i\delta$ are the roots of $x^{2}-(3-2i)x-(2i-2)=0,\ i=\sqrt{-1}$, then $\alpha\gamma+\beta\delta$ is equal to:
Step-by-Step Solution
Key Concept: Solve the quadratic with complex coefficients via the standard formula, then compute $\sqrt{-3-4i}$ by writing it as $(p+qi)^{2}$ and matching real/imaginary parts.
By the quadratic formula,
$$x=\frac{(3-2i)\pm\sqrt{(3-2i)^{2}+4(2i-2)}}{2}=\frac{(3-2i)\pm\sqrt{-3-4i}}{2}.$$
Find $\sqrt{-3-4i}$: solve $(p+qi)^{2}=-3-4i\Rightarrow p^{2}-q^{2}=-3,\ 2pq=-4$. This gives $p=1,\ q=-2$, so $\sqrt{-3-4i}=\pm(1-2i)$.
$$x_{1}=\frac{(3-2i)+(1-2i)}{2}=\frac{4-4i}{2}=2-2i,\qquad x_{2}=\frac{(3-2i)-(1-2i)}{2}=1.$$
So $(\alpha,\beta)=(2,-2)$ and $(\gamma,\delta)=(1,0)$. Therefore
$$\alpha\gamma+\beta\delta = 2\cdot 1+(-2)\cdot 0 = 2.$$
Correct Answer: 4