Showing the function $f(x,y)$ is one by one

Question

Yesterday, while teaching geometry, I was faced to a problem saying that the function below is an distance function: $$d(P,Q)=\Big|\ln\frac{\frac{x_1-c+r}{y_1}}{\frac{x_2-c+r}{y_2}}\Big|$$ where in $P|_{y_1}^{x_1}$, $Q|_{y_2}^{x_2}$ are points in $\mathbb R^2$ and $y_i>0$. In fact, I am working on Poincaré half-plane model to show that the above function has the following properties:

$*_1$ $d(P,Q)\geq 0$

$*_2$ $d(P,Q)=d(Q,P)$ for every points in above plane.

$*_3$ $d(P,Q)=0\Longleftrightarrow P=Q$.

And we don't have the triangle inequality condition here. Honestly, I chose this problem and thought it was easy to the end. Every things went OK, but while probing the second condition $(\Longrightarrow)$; I found out that I should show the function $$f(x,y)=|\ln\frac{x-c+r}{y}|$$ where $c,r$ are positive reals and $(x-c)^2+y^2=r^2$ is one by one. I could not make my guess proved in the class. Any hint would be appreciated.

Answer 1

From $(x-c)^2+y^2=r^2$ we may write $x=c+r\cos t,\ y= r \sin t.$ From $y>0$ we also have $0<t<\pi$ here. Then $$\frac{x-c+r}{y}=\frac{\cos t+1}{\sin t},$$ and the latter is strictly monotone decreasing on $(0,\pi)$.

The function $\ln$ being one to one, this shows $f(x,y)$ is also one-to-one. [note maybe the absolute values should be taken before the $\ln$ (to make input to $\ln$ positive).

Answer 2

Let $$u = x-c, y = \sqrt{r^2 - u^2}$$

Then $$\tilde{f}(x,y) = \log \frac{x-c+r}{y} = \log \frac{u+r}{\sqrt{r^2 - u^2}} = \frac{1}{2}\log \frac{r+u}{r-u} $$

Now note that the denominator decreases from $2r$ to $0$ as $u$ increases from $-r$ to $r$, and the numerator increases from $0$ to $2r$. Thus the function is monotonic for the relevant range $-r < x-c < r$, and therefore one-to-one.

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Now note that $y>0$ holds for you; and that for the condition you were trying to prove you want $\tilde{f}$, not $f = \left| \tilde{f} \right|$, to be monotonic, since $$\left| \log\frac{\cdots}{\cdots}\right| = \left| \tilde{f}(\cdots) - \tilde{f}(\cdots)\right| = 0$$ is the condition of interest.