2D仿射变换的复数公式

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2D仿射变换\[
\begin{bmatrix}x'\\y'\\1\end{bmatrix} = \begin{bmatrix}a&c&e\\b&d&f\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}\]可写成$$z' = c_1 z + c_2\bar{z} + c_3$$
其中 $z = x+iy,\ c_1 =\frac{a+d}{2} + i\frac{b-c}{2},\ c_2 = \frac{a-d}{2} + i\frac{b+c}{2},\ c_3 = e + if$.

证明

\begin{align*}
c_1z+c_2\bar{z}+c_3 &=\left( \frac{a+d}{2} +i \frac{b-c}2\right)\cdot \left(x+iy \right) + \left( \frac{a-d}{2} +i \frac{b+c}2\right)\cdot \left(x-iy \right)+c_3 \\
&= ax+idy +ibx+cy+c_3 \\
&= ax+cy + i(bx+dy)+e+if \\
&=ax+cy+e+i(bx+dy+f)\\
&= x'+iy'
\end{align*}

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Learning PostScript by Doing 6 Coordinate Transformations
如果一个点在坐标系中的坐标是 $(x, y)$，该坐标系围绕点 $(a, b)$ 逆时针旋转角度 $θ$，则其在当前页面中的坐标 $(x_\text{page}, y_\text{page})$ 为$$\left(\begin{array}{c}x_{\text {page}} \\ y_{\text {page}} \\ 1\end{array}\right)=\left(\begin{array}{ccc}\cos \theta & -\sin \theta & a(1-\cos \theta)+b \sin \theta \\ \sin \theta & \cos \theta & b(1-\cos \theta)-a \sin \theta \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{l}x \\ y \\ 1\end{array}\right)$$也写成$$z'=e^{i\theta}(z-w)+w$$其中$w=a+bi$