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.

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).