Is there a simple and **explicit** continuous function $f\colon[0,\infty)^2\to\mathbb C$ such that $f$ is analytic on $(0,\infty)^2$ and $|f(x+iy)|/(x+y)\to1$ as $x+y\to\infty$, where $(x,y)\in[0,\infty)^2$? 


It seems not too hard to construct such a function $f$ as the sum of a series of polynomials, as is done in the proof of the Carleman approximation theorem; cf. **[[1](https://mathoverflow.net/questions/26243/asymptotic-approximation-of-x-alpha-by-entire-functions/26290?noredirect=1#comment546591_26290)]**, **[[2](http://projecteuclid.org/euclid.mmj/1031710533)]**. However, that construction is far from explicit. 



Explicit analytic function with modulus asymptotic to $\Re z+\Im z$
Such function does not exist, "explicit" or not.
Consider $u(z)=\log|f(z)/z|$. Your condition implies that $u$ is harmonic and
bounded when $|z|>r$ for some $r>0$ and $z$ is in the first quadrant. Moreover it tends to $0$
on positive real and imaginary axes. But on the line $x=y$ it tends to $\log\sqrt{2}>0$. This contradicts the Phragmen--Lindelof Principle. 


EDIT. If $x=y$ then 
$$u(x+ix)=\log|f(x+ix)/(x+ix)|=\log( 2x/(x\sqrt{2}))+o(1)\to\log\sqrt{2}.$$
The form of Phragmen-Lindelof that is needed is the following: Let $u$ be harmonic in a region $D$, $u$ is bounded from above and $a\in\partial D$.
If $$\limsup\_{z\to\zeta}u(z)\leq C,$$ for all $\zeta\in\partial D\cap V$,
where $V$ is a punctured neighborhood of $a$, then
$$\limsup\_{z\to a}u(z)\leq C.$$ In our case $a=\infty$.


