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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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    @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...2012-03-07

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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 +1.2012-03-07
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$\hskip 1.3in$ rotation

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