## 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).