Converting between polar and Cartesian coordinates The polar coordinates $r$ and $\varphi$ can be converted to the Cartesian coordinates x and y 
by using the [[trigonometric function]]s sine and cosine:
$$x = r \cos \varphi \,$$
$$y = r \sin \varphi \,$$
The Cartesian coordinates ''$x$'' and ''y'' can be converted to polar coordinates ''r'' and ''$\varphi$'' with ''r'' ≥ 0 and ''$\varphi$'' in the interval $[0, 2\pi)$.
$r = \sqrt{x^2 + y^2} \quad$ (as in the (Pythagorean theorem) or the (Euclidean norm), 
$$\varphi=\begin{cases}
\arctan(\frac{y}{x}) & \mbox{if } x > 0 \mbox{and } y \geq 0\\
\arctan(\frac{y}{x}) + 2\pi & \mbox{if } x > 0 \mbox{ and } y < 0\\
\arctan(\frac{y}{x}) + \pi & \mbox{if } x < 0 \\
\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y > 0\\
-\frac{3\pi}{2} & \mbox{if } x = 0 \mbox{ and } y < 0\\
\text{undefined} & \mbox{if } x = 0 \mbox{ and } y = 0
\end{cases}$$
my question is $\varphi$ could be  written in one equation by using Congruences modulo n as follows :
$$\varphi=\arctan(\frac{y}{x}) [2\pi]$$ which means there exists some integer ”k” such that $$\varphi=\arctan(\frac{y}{x})+2\cdot k\cdot \pi $$
Can we speak of congruence modulo 2pi (for example) for angles?
Are there any serious references (if wikipedia would answer my question but I am suspicious ...) which I can support myself.
I think there is discontinuity somewhere, because when we do continuously around the origin in the direct sense, $\varphi$ angle increases by $2\pi$ and therefore can not be equal to the initial value that chose. Am i right ??

Can the angles be defined up to $2k\pi$? Yes.
Can we write $\phi = \arctan(y/x)+ 2k\pi$?
Almost, but not exactly. The problem in the use of the $\arctan$ function is not (only) that the angle is defined up to multiples of $2\pi$ (which you said correctly), but also that it gives values only between $-\pi/2$ and $+\pi/2$. 
For example, consider the point at $(-1,0)$ in Cartesian coordinates. That is, $r=1, \phi = \pi$ in polar coordinates. If we used the "function" you define, $\phi = \arctan(y/x)+ 2k\pi$, we would get:
$$
\phi = \arctan(\frac{0}{-1}) + 2k\pi = 0 + 2k\pi \ne \pi.
$$
The problem of the $\arctan$ function is precisely its codomain. Congruences can help, and are a very nice way to look at the concept of "angle", but don't solve this problem.
In a way, the definition with "multiple cases", that you wrote, is unavoidable (and it is linked to the fact that no continuous function can map the real line into the "circle" of the angles). 
In the applications (for example, in CS), the problem is solved by the "atan2" function. (The article, if you want, also answers partly to your question.)
