Amoeba of a line in the plane: An example

Question

Let $z+w+1=0$ a line in $\mathbb{C}^2$ and let $x=log|z| \ge 0$ and $y=log|w|$. I have to show that $$ log(e^x-1) \le y \le 1+e^x $$ But I can't do it! Can you help me, please?

Answer 1

$z+w+1=0\Longrightarrow w=-z-1\Longrightarrow |w|=|z-1|$.

Remember $||z|-1|\leq|z+1|\leq|z|+1$.

If $x\geq 0$ then $|z|=e^x\geq 1$ so;

$||z|=1|=|z|-1\Longrightarrow |z|-1\leq|z+1|\leq |z|+1$

$$\log(|z|-1)\leq\log |z+1|\leq\log)|z|+1)$$ $$\log(e^x-1)\leq\log |w|\leq\log(e^x+1)$$ $$\log(e^x-1)\leq\log y\leq\log(e^x+1)$$

And if $x\leq 0$ we arrive; $$\log(1-e^x)\leq\log y\leq\log(e^x+1)$$

I guess you have read it at the book Tropical Algebraic Geometry written by Itenberg, Mikhalkin and Sgustin. I guess they missed one $\log$ in that page. And as the aim is arriving a Tropical polynomial, it does need the $\log$ in the right hand side. Also the figure drawn at next page is used $\log(1+e^x)$ not $1+e^x$.