QUESTION
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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.




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(*Please, feel free to remove/add tags*).




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A connection
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(*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). 



Anything special (historical?) about surface $x\cdot y\cdot z\ +\ x+y+z=0$?
The corresponding projective surface 
$$S: xyz + (x + y + z)w^2=0 \subset \mathbb{P}^3,$$
is a singular cubic surface - singular cubic surfaces are special. It has three singularities, each of which has singularity type $A\_1$ (this means that locally each has the shape $xy + w^2 =0$). These are the points $(1:0:0:0),(0:1:0:0)$ and $(0:0:1:0)$ (note that these all lie at the "plane at infinity", $w=0$). The surface $S$ contains $12$ lines ($6$ of these are easy to find, the other $6$ take more work). A concrete reason why such surfaces are special is that the collection of projective cubic surfaces over $\mathbb{C}$ with singularity type $3A\_1$ forms a one dimensional family.


These facts (and more) can be found in the article:


Bruce and Wall - On the classification of cubic surfaces.


