Gradient of Ronkin function I have a complex curve $P(z,w)=z+w-1=0$. I get the amoeba map 
$$(z,w)\rightarrow (\log|z|,\log|w|)$$ of this curve. It's look like this http://en.wikipedia.org/wiki/Amoeba_(mathematics)  (the first picture). 
The Ronkin function is defined by
$$R(x,y)=\frac{1}{(2\pi i)^2}\int_{e^x=|z|}\int_{|w|=e^y} \log|1+z+w|\frac{dz}{z}\frac{dw}{w},$$
The gradient of it equals
$$\frac{\partial R}{\partial x}= \frac{1}{2\pi i} \int_{|w|=e^y}(\frac{1}{2\pi i} \int_{|z|=e^x} \frac{dz}{z+1+w})\frac{dw}{w}$$
It equals zero if $|z|=e^x>|1+w|$, and $\frac{\alpha}{\pi}$ by the argument principle.
Can anyone explain me that $$(\frac{\partial R(x,y)}{\partial x},\frac{\partial R(x,y)}{\partial y})$$ is map from amoeba to Newton polygone?
 A: For your particular function $P$, the gradient of the Ronkin function can be computed explicitly. There are three easy cases: If $e^x+e^y < 1$, then $\nabla R = (0,0)$; if $e^x > e^y+1$ then $\nabla R = (1,0)$; if $e^y = e^x+1$ then $\nabla R = (0,1)$. In the remaining case, there is a triangle with side lengths $(1, e^x, e^y)$. Let $\alpha$ be the angle opposite $e^x$ and $\beta$ the angle opposite $e^y$. (We can write down closed formulas using the law of cosines if we need.) Then $\nabla R = \frac{1}{\pi}(\alpha, \beta)$. Since $0 < \alpha, \beta$ and $\alpha + \beta < \pi$ (since they are two angles of a triangle), this shows that the gradient is in the Newton polytope $\mathrm{Hull}((0,0), (1,0), (0,1))$ in each case, and hits all of the Newton polytope except for the interiors of the edges.
The key facts here are the following: Let $P(z,w) \neq 0$ be any polynomial (or Laurent polynomial) with complex coefficients. Define
$$R(x,y) = \frac{1}{(2 \pi)^2} \int_{\theta=0}^{2 \pi} \int_{\phi=0}^{2 \pi} \log \left|  P(e^{x+i \theta}, e^{y+i \phi}) \right| \ d \theta d \phi.$$
This integral is absolutely convergent, even though the integrand may be $- \infty$ at some points. 
For $r \in \mathbb{R}^+$ and $w \in \mathbb{C}$, let $N(r,w)$ be the number of roots of $P(z,w)$ inside the disc $\{ |z| \leq r \}$. (Zeroes must be counted with multiplicity and, if $P$ is a Laurent polynomial, then any pole at zero must be counted with negative multiplicity.) Then, if the deriviative is defined, we have
$$\left. \frac{\partial R}{\partial x} \right|_{(x,y)}= \frac{1}{2 \pi} \int_{\phi=0}^{2 \pi} N(e^x,e^{y+i \phi}) d \phi.$$
Of course, there is a similar formula for $\partial R/\partial w$, interchanging the roles of the $x$ and $y$ coordinates.
To see this, differentiate inside the integral sign and use the fact that $\oint \frac{f'(z)}{f(z)} dz$ is always the number of zeroes of $f$ inside the contour for any analytic function to perform the integral on $\theta$. 
In your particular setting, if $e^x+e^y<1$, then there is never a zero of $z+w =1$ with $w=e^{y+i \phi}$ and $|z|\leq e^x$. So the gradient is zero in this case. A similar analysis covers the cases $e^x > e^y+1$ (there is always a zero) and $e^y > e^x+1$ (there is never a zero). The remaining case is where $(e^x, e^y, 1)$ form the sides of a triangle. In that case, with $\alpha$ as in the first paragraph, one can check that there is a zero if $\pi - \alpha < \phi < \pi+\alpha$, and no zero otherwise. So $\partial R/\partial x = \frac{1}{2 \pi} ((\pi+\alpha) - (\pi - \alpha)) = \frac{1}{\pi} \alpha$, as claimed.

In general, one can show that, if $\nabla R$ is defined, then it lies in the Newton polytope of $P$. Here is a sketch of the proof. $N(r,w)$ is always an integer between $0$ and $\deg_z P$, as it is the number of roots of a polynomial of degree $\deg_z P$ inside a certain disc. Since $\partial f/\partial x$  is an average of values of $N(r,w)$, it is likewise in $[0, \deg_z P]$. Similarly, $\partial f/\partial y$ is in $[0, \deg_w P]$. If the Newton poytope of $P$ is a rectangle, this completes the proof. 
For future purposes, if $f(z)$ is a Laurent polynomial, we define $\deg^{-} f$ to be the most negative exponent occuring in $f$ and $\deg^{+} f$ to be the most positive exponent. The corresponding bounds in this case are that $\nabla R \in [\deg_z^- P , \deg_z^+ P] \times [\deg_w^- P , \deg_w^+ P]$.
Let $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ be any matrix in $SL_2(\mathbb{Z})$. Consider the change of variables $(z,w) \mapsto (z^a w^b, z^c w^d)$, turning $P$ into a new Laurent polynomial $P' := P(z^a w^b, z^c w^d)$. One can check that the Rankin function of $P'$ is the pullback of the Rankin function of $P$  under the linear transformation $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ of $\mathbb{R}^2$. Therefore, one obtains that $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) \nabla R$ is in the rectangle $[0, \deg_z P'] \times [0, \deg_w P']$, and thus $\nabla R$ is in the parallelogram $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)^{-1} \left( [\deg^-_z P', \deg^+_z P'] \times [\deg^-_w P', \deg^+_w P'] \right)$. Intersecting these parallelograms over all $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ in $SL_2(\mathbb{Z})$ gives the result.
These results usually get cited to Passare and Rullgard, Amoebas, Monge-Ampàre measures, and triangulations of the Newton polytope. I found it a little hard to extract these specific results from the stronger ones in that paper, which is why I sketched the proofs.
A more subtle question is precisely what part of the Newton polytope is the image of $\nabla R$. In several places, I have seen the assertion that Passare and Rullgard prove that every interior point is hit. But that is not true and they don't claim to prove it. Let $(u,v)$ be in the interior of the Newton polytope. Passare and Rullgard show (proof of Theorem 4, when you unpack all the definitions) that $R(x,y) - ux - vy$ has a local minimum (in fact, $R$ is convex up, so any local minimum must be a global minimum). So, if $R$ is differentiable at that point, then $\nabla R = (u,v)$. But $R$ need not be differentiable. Indeed, if $P$ is a function only of $z$, then Jensen's formula shows that $R$ is a piecewise linear function of $x$.
I believe, but do not know a proof, that:

If none of the irreducible factors of $P(z,w)$ is a binomial, then $R$ is $C^1$.

Good references on Rankin functions are Section 2 of Mikhalkin, Amoebas of algebraic varieties (but, as noted above, Proposition 12 is false) and Rullgard's thesis.
