Let \(\alpha\) and \(\beta\) be complex numbers such that \(\alpha + \beta\) and \(i(\alpha - 2 \beta)\) are both positive real numbers. If \(\beta = 3 + 2i,\) compute \(\alpha.\)
\(\beta=3+2i\\ \alpha+\beta=k \qquad k\in R\\ so\\ \alpha=n-2i\\ \alpha-2\beta=wi \qquad w\in R\\ (n-2i)-2(3+2i)=wi\\ n-6-2i-4i=wi\\ (n-6)-6i=wi\\ n-6=0\\ n=6\\ \alpha=6-2i \)
I think I made that more complicated than necessary.
\(\alpha+\beta \in \mathbb{R} \Rightarrow Im(\alpha)+2=0 \Rightarrow Im(\alpha)=-2\\ \alpha-2\beta \in i \cdot \mathbb{R}\Rightarrow Re(\alpha) - 6 = 0 \Rightarrow Re(\alpha)=6\\ \alpha = 6-2i\)
