# Rotation and radial translation together in a group?

Being a happy beginner in algebra and groups, i would like to build a group that can systematically represent (in 2 dimensions):

1. Rotation around origin.

I have learned that complex numbers can represent rotation with some resolution $\theta$ : $$\bf M_{R(\theta)} = \begin{bmatrix}a&-b\\b&a\end{bmatrix}, \cases{a=\cos(\theta)\\b=\sin(\theta)}$$

And then translation with a resolution $k$ : $$\bf M_{T(k)}=\begin{bmatrix}1&0\\k&1\end{bmatrix}$$

Can I combine these somehow to achieve what I want? To build a general matrix with help of these matrices and their exponents $${\bf M(\theta,k)} = f({\bf M_{T(k)},M_{R(\theta)}},e_\theta,e_k)$$

Edit : An image to try and clarify : We want our group to take a given line to any other given line, we measure their distance to origo and their angle. The point of interest ( which i am talking about above ) will have a $\Delta$ in each of the dimensions: angle $\theta$ and radius translation $k$.

Example where the blue line is taken to the black line by concatenating the operations: $$\cases{r = r - 0\cdot 0.125\\\theta =\theta+ 2\cdot \frac{\pi}{8}}$$ The other tuples decide the numbers for lines of the other colors. We want to find a matrix representation of generators for a group which does this. The red helper circles are centered on origo and of radiuses $0.25,0.5$ to help see what the "translation" operation is supposed to do.

• I don't really understand what you mean by "translation along the radius" or "translation with resolution $k$"... Your matrix $\bf M_{T(k)}$ does not define a translation on $\Bbb R^2$, in fact no matrix defines a translation as translations are not linear. – Arnaud D. Feb 18 '17 at 11:23
• It is only intended as a 1D translation in the radial direction ${\bf M_{T(k)}}^n = \begin{bmatrix}1&0\\kn&1 \end{bmatrix}$ is $n$ "steps" if k is 1 "step". The length in position $(2,1)$ of the matrix. And the inverses are intended to correspond to steps in opposite direction. – mathreadler Feb 18 '17 at 11:32
• But it's not a translation, since it fixes the origin... – Arnaud D. Feb 18 '17 at 11:34
• Yes, but it is not a scaling either. I don't know the word for it. I want to traverse back and forth radially on a linear scale and let the rotation steer the angle. – mathreadler Feb 18 '17 at 11:35
• So if I understand correctly, if you have a point $P$, you want to map it to a point $Q$ that is on the same line through the origin $O$, and such that $d(O,Q)=k\cdot d(O,P)$? Or do you want something like $d(O,Q)=d(O,P)+k$? – Arnaud D. Feb 18 '17 at 12:21

Assuming that we start with closest point at the x-axis, one way to calculate the group action on a point $(x,y) \in {\mathbb R}^2$:
$$\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}\underset{\text{Rotation generator}}{\underbrace{\begin{bmatrix}a&-b&0\\b&a&0\\0&0&0\\\end{bmatrix}}}^{\,\,m}\underset{\text{Translation generator}}{\underbrace{\begin{bmatrix}1&0&k\\0&1&0\\0&0&1\end{bmatrix}}}^n\begin{bmatrix}x\\y\\1\end{bmatrix}$$
Or in vector form with $\bf V$ having columns being points to act on and $\bf t$ being a translation vector:
$$\begin{bmatrix}\bf I_2&\bf 0\end{bmatrix}\begin{bmatrix}\bf R&\bf 0\\{\bf 0}^T&0\\\end{bmatrix}^m\begin{bmatrix}{\bf I_2}&\bf t\\\bf 0&1\end{bmatrix}^n\begin{bmatrix}\bf V\\{\bf 1}^T\end{bmatrix}$$