An Attempt to Simplify Quadratic Curves Using Complex Numbers

用复数化简二次曲线的尝试 苏剑林 · 2013-01-02 · Mathematics curve complex number conic section quadratic form

An attempt to simplify quadratic curves using complex numbers

When a quadratic form is restricted to the two-dimensional plane, simplifying it becomes equivalent to simplifying a quadratic curve. This simplification mainly relies on translation and rotation — precisely the operations that complex numbers are "good at." So using complex numbers as a tool to simplify quadratic curves seems like a rather obvious idea. Yet I haven't come across any material on this, and I myself had overlooked this approach before. Below I'll explore it a little.

Since I only intend to give a heuristic illustration, I'll restrict attention here to the incomplete form $ Ax^2+2Bxy+Cy^2=1$ (which excludes the parabola). more

For a complex number $z=x+yi$, we have

$$ x=\frac{1}{2}(z+\bar{z}) \\ y=\frac{1}{2i}(z-\bar{z})$$

Substituting these two expressions gives the complex-number representation of an arbitrary planar curve. For the quadratic curve above, we obtain:

$$(A-C+Bi)z^2+(A-C-Bi)\bar{z}^2+2(A+C)z\bar{z}=4$$

For the final expression of the quadratic form, we want it to consist only of squared terms, and the complex expression for a squared term is

$$z\bar{z}=x^2+y^2 \\ z^2+\bar{z}^2=2(x^2-y^2)$$

So what we need to do is transform the complex expression of the quadratic curve into a linear combination of $z\bar{z}$ and $z^2+\bar{z}^2$. Starting directly from

$$(A-C+Bi)z^2+(A-C-Bi)\bar{z}^2+2(A+C)z\bar{z}=4$$

it's easy to see that if we set

$$Z=z\sqrt{A-C+Bi}$$

then the above expression can be rewritten as

$$Z^2+\bar{Z}^2+\frac{2(A+C)}{\sqrt{(A-C)^2+B^2}}Z\bar{Z}=4$$

which is exactly the form we wanted. What a pleasing coincidence! In this way, starting from complex numbers and going through only a little computation, we've easily arrived at the simplest form of the quadratic curve:

$$\begin{aligned}2(X^2-Y^2)+\frac{2(A+C)}{\sqrt{(A-C)^2+B^2}}(X^2+Y^2)=4 \\ (\frac{A+C}{\sqrt{(A-C)^2+B^2}}+1)X^2+(\frac{A+C}{\sqrt{(A-C)^2+B^2}}-1)Y^2=2\end{aligned}$$

Moreover, when simplifying quadratic curves using matrices, it's usually rather hard to write down the explicit form of the rotation transformation. But with complex numbers it's a different story — they tell us directly that:

$$Z=z\sqrt{A-C+Bi}$$

This shows that this kind of "geometry of numbers" can offer considerable advantages in certain cases. Although matrix algebra can fully subsume the results for higher-dimensional number systems (quaternions, octonions), studying the intrinsic patterns of the numbers themselves is still quite illuminating. It's much like how complex numbers came to occupy an essential place in modern quantum mechanics — something their inventors could never have imagined. Perhaps one day some other kind of number will play an even greater role in describing the universe. This is exactly where the charm of mathematics lies!