Proof of Pascal's Theorem (on circles) using complex numbers.

Question

I just learnt about using complex numbers in geometry and it is very useful indeed. I came to know about Pascal's Theorem (on circles) recently which is stated below:

Let $A, B, C, D, E, F$ be six distinct points on a circle $\Gamma$. Then the intersection of $AB$ and $DE$, the intersection of $BC$ and $EF$ and the intersection of $CD$ and $FA$ are collinear.

I feel like this question can be solved easily using complex numbers. Toss the circle on the complex plane as the unit circle. Get the intersections and check that these points are collinear by whatever way you may like. However, I have been stuck on multiple products and not moving any ahead. Any help would be appreciated.

Answer 1

Let's see. On a unit circle, you just have complex phases, so $$A=e^{ia}, B=e^{ib}, \ldots$$ This part is clear.

Then you have to find intersections. This part is the most involved. Let's see for $AB$ and $DE$. We have to solve $$X=A+t(B-A) = D+t'(E-D)$$ for real numbers $t$ and $t'$ which isn't any better than working with vectors. If you check this link, you will find quite a horrible formula for an intersection with complex numbers that comes from solving the above equations. It does simplify a little bit for unit circle, but not much. Let's just write it out:

$$X=2i\frac{\sin(b-a)(e^{id}-e^{ie})-\sin(e-d)(e^{ib}-e^{ia})}{(e^{-a}-e^{-b})(e^{d}-e^{e})-(e^{a}-e^{b})(e^{-d}-e^{-e})}$$

At the final step, you would need to prove collinearity, which means $XY$ and $XZ$ are real multiples of each other.

There may be a nicer complex-based solution that I can't see - looking forward to seeing another answer to point it out. Until then, I have my doubts.

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As a general advice, what simplifies and what doesn't, when you use complex numbers:

• Translations and rotations are just addition and multiplication, your results are invariant to transformations by default. Instead of matrix-vector operations you have elementary operations.
• Trigonometric identities become just algebraic rules (multiplication and addition of exponentials).
• Circles and lines are treated as equal, all line results can be mapped to circles via Möbius transformation and vice-versa.
• Laplace equation is satisfied automatically and conformal maps help you convert things into a different geometry where you can solve it.
• You have calculus - differentiation and integration with "vectors" becomes possible, usually the imaginary part of the solution gives you something for free.
• Linear differential equations with constant coefficients become trivial to solve (complex amplitudes and impedances in AC circuitry, frequency-dependent response functions, Fourier analysis,...).

What stays difficult:

• Whenever you have to convert from polar to Cartesian form, you introduce back trigonometric functions, inverse trig, Pythagorean theorem and all the nasty stuff that comes with it. That comes up every time you specifically require one number to be explicitly real (such as in our case, where we require $t$ and $t'$ to be real, and must split the equations into real and imaginary parts to solve them)
• If somehow complex conjugates become involved (for example, by having restrictions on lengths), you lose most of algebraic tools for simplification and everything just reverts back to calculating vector norms, not much gain from complex numbers. Heron's formula is one such example - it's hard to see immediately how that would be any easier to derive from complex numbers.
• Non-comformal transformations (which operate on real and imaginary components differently) for example, anisotropic stretches, making circles into ellipses, again can't be expressed with simple complex operations and you are back to doing it by components.