Check for complex chart on $\mathbb{R^2}$ 
Let $X=\mathbb{R^2}, U$ open in $X$. Define $f: U \to \mathbb{C}$ as $$f_U(x,y)= \frac{x}{1+\sqrt{x^2+y^2}} + i  \frac{y}{1+\sqrt{x^2+y^2}}$$
  I am to show that this defines a complex chart on $X$.

My solution:
I compute the norm in the complex plane as a function of the norm in $X$, $n(x) = \frac{x}{x^2+1}$ and simply check that $n(x) \to 0$ as $x \to 0$. Is this correct? 
I apologize for the poor choice of notation and lack of detail, I do not have access to a computer at the moment. I would also like to know of any quick checks that a map defines a chart.
 A: As you say, you understand a complex chart as a homeomorphism from an open subset of $\mathbb R^2 = \mathbb  C$ to an open subset of the complex plane.
We have $f_U(z) = f(z) = \dfrac{z}{1+ \lvert z \rvert}$. This map is clearly continuous. Obviuosly $\lvert f(z) \rvert = \dfrac{\lvert z \lvert }{1+ \lvert z \rvert} < 1$. Let $U(0;1)$ denote the open unit disk $\{ w \in \mathbb C \mid \lvert w \rvert < 1 \}$. Define
$$g : U(0;1) \to \mathbb C, g(w) = \dfrac{w}{1 - \lvert w \rvert} .$$
This is also a continuous map. We have
$$g(f(z)) = g(\dfrac{z}{1+ \lvert z \rvert}) = \dfrac{\dfrac{z}{1+ \lvert z \rvert}}{ 1- \dfrac{\lvert z \rvert }{1+ \lvert z \rvert}} = z$$
and
$$f(g(w))  = f(\dfrac{w}{1 - \lvert w \rvert}) = \dfrac{\dfrac{w}{1 - \lvert w \rvert}}{1 + \dfrac{\lvert w \rvert}{1 - \lvert w \rvert}} = w.$$
Thus $f$ and $g$ are inverse to each other. In other words, $f$ and $g$ are homeomorphisms such that $f^{-1} = g$.
Thus, for any open $U \subset \mathbb C$, $f_U$ maps $U$ homeomorphically onto $f(U)$ which is an open subset of $U(0;1)$ and hence an open subset of $\mathbb C$.
