inverse of the function $f(r,\theta) = (r\cos \theta, r \sin \theta)$

Question

inverse of the function $f(r,\theta) = (r\cos \theta, r \sin \theta)$

set $x = r \cos \theta$, $y = r \sin \theta$ then we have $ x^2 + y^2 = r^2$ so $r = \sqrt{x^2+y^2}$.

Now $y/x = \tan \theta$ so $ \arctan(y/x) = \theta$ for $x\not=0$. If $x=0$ we have $\theta = \text{arccot}(x/y)$ for the inverse.

In the solutions they have: in a neighbourhood of $(x_0,y_0)$ with $x_0 < 0$ we have $\theta = \arctan(y/x) + \pi$ and $\arctan(y/x) = \theta$ for $x_0 > 0$, and similarly for $y_0 >0 $and $<0$ - my question is, where did the $+\pi$ come from, and why do we distinguish if $x,y>0$ or $<0$?

Answer 1

The given function $f$ maps a pair of polar coordinates $(r, \theta)$ to Cartesian coordinates $(r \cos\theta, r \sin \theta)$.

You can fiddle with it here.

The arcus tangens function will return a value from $(-\pi/2,\pi/2)$, which covers only a total angle of $\pi$, not $2 \pi$. That is why they feature the case distinction.

$x > 0$ is for the right side half circle:

$x < 0$ for the left side half circle.

For $x = 0$ $\theta$ should be either $\pi/2$ $(y > 0)$ or $-\pi/2$ $(y< 0)$.

For $(x,y) = (0,0)$ there is no unique inverse.

Answer 2

Let a point $(x_0,y_0)\ne(0,0)$ be given, and denote by $H$ the half plane $x_0x+y_0y>0$. Then one has $$\theta=\theta_0+\arctan{x_0y-y_0x\over x_0x+y_0y}\qquad\bigl((x,y)\in H\bigr)\ ,$$ where $\theta_0$ is a polar angle of $(x_0,y_0)$.