I am working on a class project, the passage I quoted in here is from a book Complex Numbers & Geometry by Hahn, p.64.
For any four complex numbers $a$, $b$, $c$, $d$, the following identity is easy to verify:
$$(a-b)(c-d)+(a-d)(b-c) = (a-c)(b-d).$$
By triangle inequality, we obtain
$$|a-b||c-d|+|a-d||b-c| = |a-c||b-d|.$$
Now let us investigate when the inequality becomes an equality. In the case of triangle inequality, $$|z_1 + z_2| ≤ |z_1| + |z_2|,$$ equality holds iff $z_1/z_2$ is a positive real number (provided $z_1\cdot z_2 \neq 0$). Thus we are looking for a condition to ensure that $\frac{(a-b)(c-d)}{(a-d)(b-c)}$ is a positive real number.
My question is not so much about complex number but about how do you go from $(z_1/z_2)>0$ to saying that $(a-b)(c-d)/(a-d)(b-c)$ has to be also $>0$? As in elsewhere, the author tends to skip lots of detail, I think he also skips detail here.
Thank you very much for your time.
Put (the book's notation):
$$z_1:=(\alpha-\beta)(\gamma-\delta)\;,\;\;\;z_2:=(\alpha-\delta)(\beta-\gamma)$$
With this, we get that
$$|z_1+z_2|\le|z_1|+|z_2|\iff |(\alpha-\gamma)(\beta-\delta)|\le |(\alpha-\beta)(\gamma-\delta)|+|(\alpha-\delta)(\beta-\gamma)|$$
and the above is an equality iff $\;z_1z_2\neq0\;$ and
$$0<\frac{z_1}{z_2}=\frac{|(\alpha-\gamma)(\beta-\delta)|}{|(\alpha-\delta)(\beta-\gamma)|}$$