QUESTION
I wanted to introduce and develop the complex logarithm from scratch. As the result I've arrived a couple of months ago at the following identity after which the road to complex logarithm is wide open:
$$\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}\quad +\quad \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\qquad =\qquad 0$$
This holds over any field of course, and it is a parametrization of the following surface, call it $L$:
$$x\cdot y\cdot z\ +\ x+y+z\ \ =\ \ 0$$
Could you provide any references and information about this surface and the above formula. A knowledgeable friend of mine is sceptical about a geometric interest of this surface $L$. I still believe that in some ways $L$ must be interesting when it rests at the foundation of the complex logarithm.
(Please, feel free to remove/add tags).
A connection
(I mean a connection between surface $L$ and the complex logarithmic function. I'll write below just a little bit less pedantically than in a textbook for students).
Let's restrict ourselves now to the field of complex numbers $\mathbb C$. Let $a\ b\ c\in\mathbb C^*$, where $\mathbb C^* := \mathbb C\setminus \{0\}$. Then we may consider a pretty much canonical piecewise linear loop $\gamma := \overline{abca}$. When $0\notin\triangle(abc)$ (a closed solid triangle is meant) then we want to show that
$$\int_{\gamma} F = 0 $$
where $\forall_{z\in\mathbb C^*}\ F(z):=\frac 1z$. At first my goal is more modest. I want to show that when the diameter of the triangle is much smaller than $\max(|a|\ |b|\ |c|)$ (so that it already follows that $0$ does not belong to the triangle) then a crude approximation of the integral is very small. How small? Regular simplicial subdivisions of the triangle lead to about $n^2$ triangles of diameter about $\frac 1n$ (everything up to a multiplicative constant). Our integral above is a sum of about $n^2$ integrals over perimeters of all these small triangles (because the terms which come from the inside of the original triangle will cleanly cancel out--cleanly, I promise). Thus I want the crude approximations of the integrals over the perimeters of the small triangles to converge to $0$ faster than $\frac 1{n^2}$. Then the above integral indeed will be equal to $0$.
Let
$$A :=\frac{b+c}2\qquad B:=\frac{a+c}2\qquad C:=\frac{a+b}2$$
Then a crude approximation of the above integral over $\gamma$ can be defined as
$$\Lambda\ :=\ \frac{b-a}C + \frac{c-b}A + \frac{a-c}B$$
Due to the identity above we get:
$$\Lambda\ =\ \frac 14\cdot\frac{a-b}C\cdot\frac{b-c}A\cdot\frac{c-a}B$$
A similar formula holds for each small triangle of the consecutive simplicial subdivision. When the original triangle $\triangle(abc)$ is disjoint (outside) a disk of radius $r>0$, around $0$, then all respective values $A'\ B' C'$, corresponding to the small triangles, have modules greater than $r$. Thus the sum of the crude approximations will be of the magnitude about $(r\cdot n)^{-3}$. Since the number of summands is of the order $n^2$, the whole sum will be arbitrarily close to $0$.
The internal terms of the sum of the crude approximations cancel out (cleanly :-) because we have selected the mid-points of the edges of the triangles. Thus the whole sum of the crude approximations of the integrals for the triangles of a subdivision approximates arbitrarily well the original integral over $\gamma$.
(Now one can study integrals of $F(z):=\frac 1z$ over homotopic paths, etc).
