I'm reading Beardon's Algebra and Geometry.
Suppose that $zw\neq0$. Show that the segment joining $0$ to $z$ is perpendicular to the segment joining $0$ to $w$ if and only if $Re[z\bar{w}]=0$.
From here I first expanded $z\bar{w}$:
$$i \left(a_2 b_1-a_1 b_2\right)+a_1 a_2+b_1 b_2$$
Then I obtained the real part of $z\bar{w}$:
$$a_1 a_2+b_1 b_2$$
From here, I was kinda stuck on how to proceed in the demonstration but then I decided to proceed with some of the most obvious examples for getting complex numbers that wouldfit perpendicularity: $(0+7i)$ and $(7+0i)$ for which $0 \cdot 7+7\cdot0=0$ and we have one case of perpendicularity. This first case would be:
$$((a_1=0)\oplus (a_2=0))\wedge ((b_1=0)\oplus (b_2=0))$$
With this reasoning in mind, I thought that for all the other cases of perpendicularity, we must find any numbers that satisfy $a_1 a_2=n$ and $b_1 b_2=-n$.
Is my reasoning correct? Is there something I should add?