When a quadratic form is in the case of a two-dimensional plane, it is equivalent to the simplification of a quadratic curve. The simplification of quadratic curves mainly involves translation and rotation, which are exactly what complex numbers are “good at.” Therefore, using complex numbers as a tool to simplify quadratic curves seems like a very obvious idea. However, I have not seen much content on this, and I had previously overlooked this approach myself. Below, I will explore this idea a bit.
Since I only intend to provide some heuristic guidance, I will only consider the incomplete form Ax^2+2Bxy+Cy^2=1 (which does not include parabolas).
For a complex number z=x+yi, we have: x=\frac{1}{2}(z+\bar{z}), \quad y=\frac{1}{2i}(z-\bar{z}) By substituting these two expressions, we can obtain the complex expression for any plane curve. For the aforementioned quadratic curve, we get: (A-C+Bi)z^2+(A-C-Bi)\bar{z}^2+2(A+C)z\bar{z}=4
For the final expression of the quadratic form, we hope to have only squared terms. The complex expressions for the squared terms are: z\bar{z}=x^2+y^2, \quad z^2+\bar{z}^2=2(x^2-y^2)
Therefore, what we need to do is to transform the complex expression of the quadratic curve into a linear combination of z\bar{z} and z^2+\bar{z}^2. From the expression: (A-C+Bi)z^2+(A-C-Bi)\bar{z}^2+2(A+C)z\bar{z}=4 it is easy to see that if we let: Z=z\sqrt{A-C+Bi} the above equation can be rewritten as: Z^2+\bar{Z}^2+\frac{2(A+C)}{\sqrt{(A-C)^2+B^2}}Z\bar{Z}=4
This is the form we expected. This is truly a pleasant coincidence! Thus, starting from complex numbers and performing a bit of calculation, we easily arrive at the simplest form of the quadratic curve: \begin{aligned} 2(X^2-Y^2)+\frac{2(A+C)}{\sqrt{(A-C)^2+B^2}}(X^2+Y^2)=4 \\ \left(\frac{A+C}{\sqrt{(A-C)^2+B^2}}+1\right)X^2+\left(\frac{A+C}{\sqrt{(A-C)^2+B^2}}-1\right)Y^2=2 \end{aligned}
Furthermore, when using matrices to simplify quadratic curves, the specific form of the rotation transformation is relatively difficult to provide, but complex numbers are different; they tell us directly: Z=z\sqrt{A-C+Bi}
It can be seen that this kind of geometry of numbers has considerable advantages in certain cases. Although matrix algebra can completely encompass the results of higher-dimensional numbers (quaternions, octonions), studying the laws of the numbers themselves is still quite enlightening. Just as complex numbers occupy a fundamental position in modern quantum mechanics, which was completely unimaginable to their inventors, perhaps one day, a certain type of number will play an even greater role in describing the universe. This is the charm of mathematics!
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