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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 ??

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  • $\begingroup$ yes. it can.${}$ $\endgroup$ Commented Mar 16, 2014 at 17:06
  • $\begingroup$ Serious references? Any trigonometry section of a pre-calculus textbook... "Angle" for trigonometric functions is taken to be any real number, and values $ \ 2 \pi \ $ apart are taken to be different "names" for the same direction. $\endgroup$ Commented Mar 16, 2014 at 17:32
  • $\begingroup$ @RecklessReckoner name good one please ! $\endgroup$ Commented Mar 16, 2014 at 17:37
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    $\begingroup$ Michael Sullivan, Pre-calculus (9th edition) is a reasonable choice. But there are dozens of such books and numerous reliable on-line sources as well. This fact doesn't really require a citation. $\endgroup$ Commented Mar 16, 2014 at 17:40
  • $\begingroup$ @RecklessReckoner Thanks $\endgroup$ Commented Mar 16, 2014 at 17:44

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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.)

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  • $\begingroup$ if x = 0 and y<0, OP has written $phi = -3 \pi/2$. Is this correct in the range [0, 2pi)? $\endgroup$ Commented Oct 30, 2024 at 14:59
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Speaking of congruence mod $2\pi$ makes perfectly good sense for real numbers (define an equivalence relation on $\mathbf{R}$ by $x \sim y$ if and only if $y - x = 2\pi k$ for some integer $k$). However, you cannot use congruence mod $2\pi$ to define the polar angle by a single formula: $\arctan(y/x) = k$ along the line $y = kx$ (with the origin removed), but the polar angle $\theta$ differs by $\pi$ on the two rays.

Incidentally, with your piecewise definition of $\varphi$ it's not easy to see $\varphi$ is a smooth function on the plane with a closed ray removed. Instead, you can use $$ \varphi = \begin{cases} \operatorname{arccot}(x/y) - \pi & \text{if $y < 0$,} \\ \arctan(y/x) & \text{if $x > 0$,} \\ \operatorname{arccot}(x/y) & \text{if $y > 0$,} \\ \end{cases} $$ and then show that the first two formulas agree in the open fourth quadrant, while the last two agree in the open first quadrant. (This defines $\varphi$ as a smooth function from the slit plane with the non-positive $x$-axis removed to the interval $(-\pi, \pi)$.)

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