As shown in the figure, the coordinate (x, y) is rotated counterclockwise by an angle \theta around the point (p, q) to obtain the coordinate (x', y'). Find the expressions for x' and y' in terms of x and y.
We have discussed this problem before. In the article "Function Image Rotation Formula", we analyzed it using methods of analytical geometry. That article was completed in February, at a time when I had not yet systematically studied vectors and complex numbers. Now, BoJone will provide two proofs of the rotation formula from the perspectives of vectors and complex numbers for reference. If there are any errors, please point them out.
To simplify the problem, we first perform the following translation:
In this way, we only need to discuss the case where the center of rotation is at the origin. First, we use vectors to solve it. The two points before and after rotation are represented as vectors \vec{A}=(x-p, y-q, 0) and \vec{B}=(x'-p, y'-q, 0), with |\vec{A}|=|\vec{B}|=R. Then we have \vec{A} \times \vec{B} = (0, 0, (x-p)(y'-q)-(x'-p)(y-q)), and: \vec{A} \cdot \vec{B} = R^2 \cos\theta = (x-p)(x'-p) + (y-q)(y'-q) \tag{1} |\vec{A} \times \vec{B}| = |R^2 \sin\theta| = |(x-p)(y'-q) - (x'-p)(y-q)| \tag{2}
Considering the case where 0 < \theta < \pi:
(1) \times (y-q) + (2) \times (x-p) yields: \begin{aligned} (y'-q)[(y-q)^2 + (x-p)^2] &= R^2[(y-q)\cos\theta + (x-p)\sin\theta] \\ y'-q &= (y-q)\cos\theta + (x-p)\sin\theta \end{aligned} \tag{3}
(1) \times (x-p) + (2) \times (y-q) yields: \begin{aligned} (x'-p)[(y-q)^2 + (x-p)^2] &= R^2[(x-p)\cos\theta - (y-q)\sin\theta] \\ x'-p &= (x-p)\cos\theta - (y-q)\sin\theta \end{aligned} \tag{4}
Equations (3) and (4) are the coordinate rotation formulas. The forms are the same when \pi < \theta < 2\pi.
Next, we use complex numbers to solve it. We know that complex numbers can be represented on the complex plane, and when two complex numbers are multiplied, the result is also a complex number whose magnitude is the product of the magnitudes of the factors, and whose argument is the sum of the arguments of the factors. Thus, we might as well use z_1 = (x-p) + (y-q)i to represent the point before rotation, and z_2 = (x'-p) + (y'-q)i to represent the point after rotation. Clearly, z_2 is obtained by multiplying z_1 by a complex number with a magnitude of 1 and an argument of \theta. It is not difficult to see that this complex number is \cos\theta + i \sin\theta. That is:
\begin{aligned} [(x-p) + (y-q)i] \cdot [\cos\theta + (\sin\theta)i] &= (x'-p) + (y'-q)i \\ [(x-p)\cos\theta - (y-q)\sin\theta] + [(y-q)\cos\theta + (x-p)\sin\theta]i &= (x'-p) + (y'-q)i \end{aligned}
According to the condition for the equality of complex numbers: \begin{aligned} y'-q &= (y-q)\cos\theta + (x-p)\sin\theta \\ x'-p &= (x-p)\cos\theta - (y-q)\sin\theta \end{aligned}
A very important point in using complex numbers to solve geometric problems is applying the condition that complex numbers are equal if and only if "real part = real part and imaginary part = imaginary part." This sometimes allows the problem to be brought back into the realm of real numbers, which can then be solved using known knowledge. To better apply complex numbers to geometry, one must also be familiar with the application of the complex plane, the key being to understand the geometric meaning of complex number operations.
Finally, let’s look at a derivation using polar coordinates. From the figure above, we can see: x-p = r \cos f, \quad y-q = r \sin f, \quad r = \sqrt{(x-p)^2 + (y-q)^2} And we have: \begin{aligned} x'-p &= r \cos (f+\theta) = r \cos f \cos\theta - r \sin f \sin\theta = (x-p)\cos\theta - (y-q)\sin\theta \\ y'-q &= r \sin (f+\theta) = r \sin f \cos\theta + r \cos f \sin\theta = (y-q)\cos\theta + (x-p)\sin\theta \end{aligned}
Q.E.D.
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