Showing that the segment joining $0$ to $z$ is perpendicular to the segment joining $0$ and $w$ iff $Re[z\bar{w}]=0$

Question

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?

The vectors $(a_1,b_1)$ and $(a_2,b_2)$ are perpendicular, or orthogonal, if their dot product, or inner product, is zero. In $\mathbb{R}^2$, that is $a_1a_2+b_1b_2=0$. In particular, see the end of this section. — robjohn, May 15 '13 at 20:45

score 2 · Accepted Answer · answered May 14 '13 at 08:28

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Two lines are perpendicular iff the scalar product of some vectors in their direction is zero. Apply this to the vectors $(a_1,b_1)$ and $(a_2,b_2)$.

answered May 14 '13 at 08:28

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Showing that the segment joining $0$ to $z$ is perpendicular to the segment joining $0$ and $w$ iff $Re[z\bar{w}]=0$

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