Find the inverse of $\frac{x}{\|x\|}$ in $\mathbb{R^2}$

Question

I wish to find the inverse of $\dfrac{x}{\|x\|}$, where $x \in \mathbb{R}^2$

Let's do this.

Let $$y_1 = \dfrac{x_1}{\sqrt{x_1^2+x_2^2}}$$

$$y_2 = \dfrac{x_2}{\sqrt{x_1^2+x_2^2}}$$

Then $$y_1 = \dfrac{x_1}{\sqrt{x_1^2+x_2^2}} \Rightarrow \dfrac{x_2^2}{x_1^2} = \dfrac{1 - y_1^2}{y_1^2}$$

Similarly

$$y_2 = \dfrac{x_2}{\sqrt{x_1^2+x_2^2}} \Rightarrow \dfrac{x_2^2}{x_1^2} = \dfrac{y_2^2}{1-y_2^2}$$

It seems however we combine the above equations, the $x_1$ and the $x_2$ will both drop out. Leaving only the $y$s

How to proceed from here?

Answer 1

The answer is quite easy if you think of what is going on with the transformation. We are taking a vector of $\mathbb{R}^2$ and assigning to it the unique unit vector which has the same direction and orientation.

So the inverse of each unit vector $u$ is the ray which goes from 0 to u (without including the 0).

Note however that there is not an inverse function, since the mapping is not one-to-one.

Answer 2

Let $x_1 = cos \theta$, $x_2 = sin \theta$, where $\theta = arctan(\frac{x_2}{x1})$, then $$y_1 = cos(arctan(\frac{x_2}{x_1})$$ $$y_2 = sin(arctan(\frac{x_2}{x_1})$$