Segment by $a$ and $b$ and let $z$ be a complex number not in it. Show that $\frac{z-a}{z-b}$ isn't a real $\le 0$ I need to show that, in a segment $ab$, when $z$ is a complex number outside it, the expression below is never a real $\le 0$.
$$\frac{z-a}{z-b}$$
I think it has something to do with the Log function, as it's not analytic exactly in the real negative line with $0$ included.
Later in this exercise it asks me to derivate $\log \frac{z-a}{z-b}$ and I did:
$\log \frac{z-a}{z-b} = \log (z-a)-\log (z-b)$ then I took the derivative, but I need to know that the expression above is never on the negative real axis.
Any ideas?
 A: This doesn't need complex logarithms or anything advanced — just algebra.
First notice that the line segment from $\,a\,$ to $\,b\,$ is the set of all points of the form $\;a+(b-a)t\;$ where $\,t\,$ is a real number in $\,[0,1].$
Now, if $\;\dfrac{z-a}{z-b}=-r,\;$ where $r$ is a non-negative real, then solving for $\,z\,$ yields $$z=\dfrac{a+rb}{1+r}$$ (note that the denominator can't be $0,$ since $r$ is non-negative).
It follows that
\begin{align}\require{cancel}z&=\frac{a}{1+r}+\frac{ra}{1+r}+\frac{r(b-a)}{1+r} \scriptsize{\quad(\text{adding and subtracting }\frac{ra}{1+r})}
\\&=\frac{a(1+r)}{1+r}+\frac{r}{1+r}(b-a)
\\&= a+\frac{r}{1+r}(b-a)
\end{align}.
But $\;\dfrac{r}{1+r}\;$ is a real number in $[0,1],$ so $\,z\,$ must be on the line segment connecting $\,a\,$ and $\,b.$
A: $M(z)=\frac{z-a}{z-b}$ is a Mobius transformation so it is a bijection of the extended complex plane $\mathbb{C}\cup\{\infty_\mathbb{C}\}$. This says that it is equivalent to show that
$$
M([a,b])=\mathbb{R}_{\leq0}\cup\{\infty_\mathbb{C}\}.
$$
What are the values of $M(a+tb)$ for $t\in[0,1]$?
Edit: If you're not comfortable with Mobius transformations notice that you just need to prove that $M(z)$ is injective and then that $M([a,b))=\mathbb{R}_{\leq0}$.
A: This is equivalent of showing that $$arg(\frac{z-a}{z-b}) \neq \pi$$
Thinking of $v=z-a$ and $w=z-b$ as vectors, this will only happen if $v$ and $w$ are  parallel vectors with opposite directions, and it is trivial to see that this will only happen if $z$ is in the segment between $a$ and $b$.
