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When inverting the polar coordinates, one needs the arc tan function $\arctan$ with two arguments. If
 , then $ \arctan(x,y) $ is defined as the angle $(x,y)$ makes with the positive $x$ -axis.
One usually sees expressions like $\arctan(y/x)$ , which is equal to $\arctan(x,y)$ when $(x,y)$ is in the first quadrant. However, $\arctan(y/x)$ does not give the correct angle when $(x,y)$ is in the third quadrant (since $y/x=(-y)/(-x)$ ). Also, the quotient $y/x$ involves a division by zero when $x=0$ , which is damaging both numerically and mathematically.
In most mathematical software and programming languages the two-argument $\arctan$ is directly implemented.
In Python language the functions atan(x) and atan2(x,y) are the respective one and two argument versions of $\arctan$ . The point of having the two argument version is to determine the correct quadrant of the point. For instance, $1/1 = 1 = -1/-1$ , so atan(x) cannot distinguish between $(1,1)$ and $(-1,-1)$ , but atan2(x,y) can, as the following Python code illustrates:
>>> from math import *
>>> print atan(1)
0.785398163397
>>> print atan2(1,1)
0.785398163397
>>> print atan2(-1,-1)
-2.3619449019
because $(1,1)$ has argument $\pi/4=0.7853\ldots$ but $(-1,-1)$ has argument $-3\pi/4=-2.3619\ldots$ .
In mathematical works, $\arctan(x,y)$ is simply denoted by $\theta(x, y)$ . The symbol $\theta$ obviously refers to the angle, but it is really the function $h_2$ , where $$ g(r, \theta) = (r \cos \theta, r \sin \theta)\,, \quad h(x,y) = g^{-1}(x, y) = (r, \theta)\,. $$ The function
 is the polar-to-Cartesian coordinate transformation. By the inverse function theorem, the function $h$ (the Cartesian-to-polar coordinate transformation) exists and is smooth wherever it is defined. Note that $h$ cannot be defined continuously everywhere, because of the multi-valued nature of $\theta$ -- $(r, \theta)$ and $(r, \theta + 2\pi n)$ always map to the same point under $g$ . (Similarly, $\theta$ cannot defined when $r = \sqrt{x^2 + y^2} = 0$ .) This means, if one chases a loop (say a circle) around the origin, $\theta$ would move from $0$ to $2\pi$ , even though the image point $g(r, \theta)$ winds back to the starting point.
Technically, a ``largest'' possible domain of $h$ (and $\theta$ ) can only be taken to be some simply connected open subset of
 . (Note:
 itself is not simply connected.) For example, such a domain might be
 , i.e. delete the negative real axis from
 .
The exterior derivative of $\theta$ is $$ d\theta = \frac{-y}{x^2 + y^2} \, dx + \frac{x}{x^2 + y^2} \, dy \,, $$ (found by implicit differentiation), and hence $$ \frac{\partial \theta}{\partial x} = \frac{-y}{x^2 + y^2}\,, \quad \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2} $$ (which can also be found by differentiating $\arctan(y/x)$ directly and piecing the results for each quadrant).
Of course, the formulas above are only valid wherever $\theta$ is defined, but the analytical expressions do not change no matter which domain of definition is taken for $\theta$ . This allows for the following neat formula to find the total variation of angle of a smooth curve
![$ \gamma \colon [a, b] \to \mathbbmss{R}^2 \setminus \{0 \}$](http://images.planetmath.org:8080/cache/objects/7105/js/img7.png "$ \gamma \colon [a, b] \to \mathbbmss{R}^2 \setminus \{0 \}$") : $$ \int_{\gamma} d\theta = \int_a^b \gamma^* d\theta = \int_a^b \left( \frac{-y \dot{x} }{x^2 + y^2} + \frac{x \dot{y}}{x^2 + y^2} \right) \, dt\,. $$ (This is related to the formula for the winding number and the argument principle in complex analysis.)
For example, if $\gamma(t) = (r \cos t, r \sin t)$ , for $t \in [0, 2\pi n]$ , is the circle that winds around the origin $n$ times, then $\int_{\gamma} d\theta = 2\pi n$ .
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