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What are some proofs of the Pythagorean theorem that use imaginary numbers?

There seems to be a lack of such proofs found on the internet, hence my question.

I will post my attempt, as an answer below.

Dan
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  • I'll put the gist of my comments under your post in the main comment section. A real proof would require a notion of length formally defined independently of the Euclidean metric, since the Euclidean metric makes the Pythagoras theorem definitionally true for the standard orthogonal coordinate axes and then true by basic linear algebra for a generic pair of orthogonal lines. I think the important thing is that Pythagoras' theorem has proofs, or perhaps "proofs", independent of $\Bbb R^2$ in some more primitive geometric framework, so we define distance in $\Bbb R^2$ in the Pythagorean way – FShrike Dec 10 '24 at 23:59
  • (cont.) with the intent of having a successful theory of $\Bbb R^2$ as a geometric object. But this is human fuzz and intuition. In a way, I think trying to prove Pythagoras' theorem using $\Bbb R^2$ or $\Bbb C$ is immoral; proving it here is finnicky, perhaps not quite possible in any interesting way, and not what I consider to be the real point. My two cents – FShrike Dec 11 '24 at 00:00
  • @FShrike Here is an (alleged) proof that is based on a rectangular coordinate system (but does not involve imaginary numbers). Would you say that this is a valid proof? – Dan Dec 11 '24 at 07:35
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    It presupposes the circle is carved out as the level set of a suitably nice function $\Bbb R^2\to\Bbb R$. Proofs like these are really nice (though attributing it to this one person doesn't make sense, since this observation has been made countless times) but to really nail it down formally we have to (1) take lengths valued in $\Bbb R$ and (2) have that the circle is, basically, a one dimensional real manifold (or something equivalent) which I challenge someone to state, never mind prove, using only a nebulous Euclidean notion of length (and not an actually defined metric) – FShrike Dec 11 '24 at 13:00
  • So, great; it affirms that calculus notions of tangent are doing the right thing. When I first noticed this years ago, it made me smile - "hey, geometry works". But I still stand by my previous comments – FShrike Dec 11 '24 at 13:02
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    Posted at MO. – Dan Dec 15 '24 at 07:48
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    Incidentally, we can generate Pythagorean triads using complex numbers. Let $z=u+vi$, for integers $u,v$. Let $w=z^2=(u^2-v^2)+(2uv)i$ Then $$|w|^2=(u^2+v^2)^2=(u^2-v^2)^2+(2uv)^2$$ – PM 2Ring Dec 16 '24 at 07:10
  • I’m voting to close this question because it was cross-posted. – Shaun Jan 28 '25 at 20:26

1 Answers1

2

(Self-answering)

Here is my attempt at such a proof, but I'm not sure if my proof is free of circular reasoning (does it assume the Pythagorean theorem somewhere?).

For every complex number $a+bi$, where $a,b\in\mathbb{R}$, there is associated with it a (possibly degenerate) right triangle with hypotenuse $r$ and angle $\theta$, as shown in the following Argand diagram, in which the axes are perpendicular.

enter image description here

$\begin{align} a^2+b^2&=(a+bi)(a-bi)\\ &=r(\cos\theta+i\sin\theta)r(\cos(-\theta)+i\sin(-\theta))\\ &=re^{\theta i}re^{-\theta i}\\ &=r^2\\ \end{align}$

Remarks:

  • The functions cosine and sine can be defined in terms of a unit circle, without assuming that the equation of the unit circle is $x^2+y^2=1$.
  • The link between the modulus-argument form, and the exponential form, can be proved without using the Pythagorean theorem, for example here.
  • Here is an almost identical proof, which received some objections; the difference is that, in my proof, I define the Argand diagram to have perpendicular axes.
Dan
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  • How do you define the unit circle, in your first bullet point? – FShrike Dec 10 '24 at 22:33
  • @FShrike The unit circle can be defined as the set of points in the plane that are one unit distance from the origin. – Dan Dec 10 '24 at 22:47
  • But how are you defining your metric? This is key. – FShrike Dec 10 '24 at 22:55
  • So, given complex numbers $(a,b)$, we must define their distance from the origin in such a way as to make the statement of Pythagoras' theorem actually interesting and obviously connected with "real geometric length". I believe this cannot be done; I certainly have never seen it done – FShrike Dec 10 '24 at 23:04
  • This is my major objection with attempts at proving Pythagoras. Either your metric is already defined so that Pythagoras' theorem is correct (by default) or your challenge is to relate some Euclidean primitive notions of length with some metric function $d:\Bbb R^2\times\Bbb R^2\to\Bbb R$ – FShrike Dec 10 '24 at 23:05
  • @FShrike Can we define the distance between two points in the complex plane (i.e. the origin and the point associated with $a+bi$) by just taking a ruler and measuring the distance? I don't mean to be facetious or anything like that; I suspect that I may be failling to grasp some subtlety. – Dan Dec 10 '24 at 23:15
  • I philosophically object to that. I certainly mathematically object to it, since you have not defined your ruler. But really, I see Pythagoras' theorem in the complex plane as definitionally true. The human-arbitrary choice of drawing Argand diagrams with orthogonal real and imaginary axes and the human-arbitrary choice of Euclidean metric just happens to very nicely marry the algebra of complex numbers with nice geometric ideas, so it is a successful choice that we have made, but it is a choice that is not forced. – FShrike Dec 10 '24 at 23:55
  • Because Pythagoras' theorem has strong geometric motivations / proofs in the old Euclidean framework, we have 'ported it' to $\Bbb R^2$ definitionally, because this is the most reasonable thing to do. And it works – FShrike Dec 10 '24 at 23:56
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    @FShrike FYI, I've posted this question at MO. – Dan Dec 15 '24 at 07:49