Motivation for the study of amoebas.

I heard they came from studying Laurent series in several variables. Consider Pascal's triangle of $\binom{n}{k}$ for $n \geq k \geq 0$. Can we find a generating series? Perhaps

$$ \sum_{n \geq k \geq 0} \binom{n}{k} x^k y^n = \sum_{n \geq 0}(x+1)^n y^n = \frac{1}{1 - (x+1)y}$$

This suggest maybe trying an alternative that makes $x$ and $y$ more symmetric:

$$ \sum_{n \geq k \geq 0} \binom{n}{k} x^k y^{n-k} = \sum_{n \geq 0}(x+y)^n = \frac{1}{1 - (x+y)}$$

This only works if $|x + y| < 1$, since we get convergence. Probably we also need $|x| < 1$ and $|y| < 1$ for the first sum to work.

Using Cauchy integral formula, we can recover Laurent series from this rational function. So, if $f(z) = a_0 + a_1 z + a_2 z^2 + \dots $ then

$$ a_n = \oint \frac{dz}{2\pi i z} \times z^{n+1}f(z)$$

Could this formula work in two dimensions. Maybe we can recover the binomial coefficients like so:

$$ \binom{m+n}{m} = \oint \frac{dz}{2\pi i z} \oint \frac{dz}{2\pi i z} \cdot z^{m+1}w^{n+1} \cdot \frac{1}{1- (z+w)}$$

Here $\times, \cdot$ are just multiplication over complex numbers.

This is a double contour integral and whether the "pole" $z+w =1$ lies inside the contour is related to amoebas, I think.

Then you have to solve the equation $|z+w| = |e^{2\pi i \theta} + e^{2\pi i \phi}|\leq 1$ and these are related to the sides of the triangle.

arXiv:math/0701039 - How to compute $\sum \tfrac{1}{n^2}$ by solving triangles (Mikael Passare)

EDIT I found the statement in Discrimants, Resultants and Multidimensional Determinants by Kapranov, Fomin and Zelevinsky.

Chapter 6, Corollary 1.6 Let $f(x)$ be a Laurent polynomial. All the components of the complement $\mathbb{R}^k - \log (Z_f)$ to the amoeba of $f$ are convex subsets in $\mathbb{R}^k$. They are in bijective correspondence with the Laurent series expansions of the rational function $\tfrac{1}{f(x)}$.

That is what I did here, is work through different expansions of $\tfrac{1}{1-(x+y)}$ and understand why the equation amoeba $\{ (\log|z|, \log|w|) : z+w = 1 \}$ makes an appearance.