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I know that rotation can be understood by simple complex transformation (as shown on 758)

$$\begin{align*}y_{1}+iy_2 &= \left( \cos(\alpha) + i \sin(\alpha) \right) \left( x_{1}+ix_{2} \right) \\ &= \left( x_{1}\cos(\alpha)-x_{2}\sin(\alpha) \right) +i \left( x_{1} \sin(\alpha)+x_{2}\cos(\alpha)\right) \end{align*}$$ which gets us:

$$A = \begin{pmatrix}\cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \\\end{pmatrix}$$

now I have a problem, p.762 in the book (not English), where it asks a rotation around a point $(2,3)$. I can solve this if I can understand what it means "about a point (2,3)"?

Initially, I thought that it means just a translation $b:(a,b)\rightarrow (2,3)$ but it looks a bit bizarre because for arbitrary point, the trasformation would become $B (2,3)+b^{-1}$ where $b^{-1}:(x_{1},x_{2})\mapsto(x_{1}+(a-2),x_{2}+(b-3))$ i.e. the translation back (I may have messed up with some minus there).

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Given a point $x_1$ and $x_0$, rotating $x_1$ around $x_0$ $\alpha$ degrees gives a new point $x_1'=A(\alpha)(x_1-x_0)+x_0$. Is this what you are asking? –  Shiyu Mar 7 '12 at 5:27
    
Um ... imagine having a coordinate system and some figure drawn on a sheet of tracing paper lying atop it. Stick a thumbtack through the tracing paper at point (2,3) in the coordinate system and fix it to the coordinate board behind it. You can then turn (but not move) the tracing paper in such a way that the thumbtack stays where it is. This produces a rotation about (2,3) of the figure. –  Henning Makholm Mar 7 '12 at 5:28
    
Rotations are not generally commutative with translations. –  anon Mar 7 '12 at 5:30
    
@anon: yes, $x_{1}'=A(\alpha)(x_{1}-x_{0})+x_{0}$ made be think the same. Perhaps, it is just that dead simple with the $x_{0}$, thinking...around a point $x_{0}$ a rotation, trying to visualize... –  hhh Mar 7 '12 at 5:34

2 Answers 2

up vote 1 down vote accepted

Since you know how to do rotations about the origin, first move the point $(2,3)$ to the origin, then do the rotation, and finally move the origin back to $(2,3)$.

Rotations don't generally commute with translations (if they did, what I describe above would just result on a rotation about the origin): consider the translation $(x,y)\mapsto(x+1,y)$, and the rotation by 180 degrees about the origin. If we translate first and then rotate, then $(0,0)$ goes to $(1,0)$ and then to $(-1,0)$; but if we rotate first and then translate, then $(0,0)$ goes to $(0,0)$ and then $(1,0)$.

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...clever, yes that makes it actually dead easy, thanks +1. –  hhh Mar 7 '12 at 5:46

$\hskip 1.3in$ rotation

$$\large y=\underbrace{\underbrace{A(\underbrace{x-p}_{\text{translate}})}_{\text{rotate}}+p}_{\text{translate back}} $$

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