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So basically this is a very simple algebraic proof of Pythagoras theorem, but I never saw it anywhere so I'm wondering if this is valid (or already presupposes the pythagoras theorem).

Inspired by $a^2-b^2=(a+b)(a-b)$ you can write the following:

$$ a^2+b^2 = (a+bi)(a-bi)$$

$$ a+bi = c e^{i \cdot \theta} $$

$$ a-bi= c e^{i\cdot-\theta} $$

$$ (a+bi)(a-bi) = \left( c e^{i\cdot\theta} \cdot c e^{i\cdot-\theta} \right)$$ $$ (a+bi)(a-bi)= c^2 e^{i(\theta-\theta)}=c^2 e^{i\cdot0} = c^2 \cdot 1$$ $$ a^2+b^2 = c^2 $$

If it is not valid, how could you make it valid? source https://oer.physics.manchester.ac.uk/Math2/Notes/Notes/argand.png

bananenheld
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    To make it valid, you need to connect $a$, $b$, and $c$ to sides of a right triangle somehow. When you do this, you have to make sure you're not already implictly using the Pythagorean theorem. – Jason DeVito - on hiatus Feb 17 '23 at 15:51
  • But isn't this right triangle implied by the complex number/complex plane ( $c$ being the radius with angle $\theta$ of the real number line. – bananenheld Feb 17 '23 at 15:52
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    It could be, but none of that is included in your proof. All I'm really getting at is that the Pythagorean theorem is a statement about sides of a right triangle. If your proof doesn't mention sides of a right triangle anywhere, something must be missing. – Jason DeVito - on hiatus Feb 17 '23 at 16:01
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    I agree with @JasonDeVito, this only proves that sum of squares of two reals is equal to a square of anther real. – Vasili Feb 17 '23 at 16:06
  • I added an image of the complex plane. – bananenheld Feb 17 '23 at 16:07
  • Have a look at the various proofs listed here, for example. The one using calculus might interest you, too. – Dietrich Burde Feb 17 '23 at 16:10
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    Since the notion of Euclidean distance takes Pythagoras theorem as a given (in a sense), it's difficult to prove the theorem in this way. This is because you need the notion of Euclidean distance to define the polar form of complex numbers. – Enforce Feb 17 '23 at 16:21
  • @Enforce, could you elaborate that? Why do you need the theorem to define the polar form of complex numbers, can't you just use a infinite sum to connect $a+bi$ to $e^{i\theta}$? – bananenheld Feb 17 '23 at 20:11
  • Connecting $a+bi=ce^{i\theta}$ doesn't require pythagoras. Saying that $c$ is the correct "length", however, does @bananenheld (or at least, any method that shows $c$ is the correct length is a method that basically has already proven / can easily prove pythagoras' theorem) – FShrike Feb 19 '23 at 15:27

3 Answers3

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Your arguement can be made right.

We start by defining sine and cosine as the legs of a right triangle with $\theta$ radians being the anticlockwise angle between the $x-$axis and the other side subtending the circle. Define $\cos\theta$ as the $y-$coordinate of the point at which the subtending line intersects the unit circle and $\sin\theta$ as the $x-$coordinate of the same point.

This was lengthy, but it's precise enough to draw a picture.

Now, we aim to prove/deduce the derivatives without using the pythagorean theorem.

We can prove the angle-sum identities by multiplying rotation matrices for example (the "fastest" way I could think of). $$\sin(x+y)=\sin x\cos y+\sin y\cos x\\ \cos(x+y)=\cos x\cos y-\sin x\sin y$$ We start with $\sin\theta.$ $$\sin'\theta=\lim_{h\to0}\frac{\sin(\theta+h)-\sin\theta}{h}\\ =\lim_{h\to0}\frac{\sin\theta\cos h+\sin h\cos\theta-\sin\theta}{h}\\ =\sin\theta\,\underbrace{\lim_{h\to0}\frac{\cos h-1}{h}}_{\cos'(0)}+\cos\theta\,\underbrace{\lim_{h\to0}\frac{\sin h}{h}}_{\sin'(0)}\\ =\cos\theta\sin'(0)\\ =\cos\theta$$ where $\cos'(0)=0$ because $\cos(0)=1$ is a maximum since $\cos\theta$ is bounded by a maximum of $1$ (the length of the hypotenuse $\equiv$ the radius of the unit circle) and $\sin'(0)=1$ is a classic result that can be shown by the squeeze theorem.

We can similarly prove that $\cos'\theta=-\sin\theta.$

Now, we prove Euler's identity.

Consider the function $$f(\theta)=(\cos\theta+i\sin\theta)e^{-i\theta}\\ f'(\theta)=(-\sin\theta+i\cos\theta-i\cos\theta+\sin\theta)e^{-i\theta}=0\\ f(\theta)=C\\ f(0)=1\\ f(\theta)=1\\ e^{i\theta}=\cos\theta+i\sin\theta.$$

Thus, we can prove the pythagorean identity for any right triangle on the unit circle using your arguement and scale the real and imaginary parts, effectively scaling the sides of the triangle, by a factor of $c$ so that the arguement encompasses all right triangles.$\tag*{$\blacksquare$}$

user7777777
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  • I'm not sure I understand your proof of Euler's identity. If you're proving it, then you must already have a definition of $e^{-i\theta}$. So, what is it? Without such a definition, you can't be certain that $(e^{-i\theta})' = -ie^{-i\theta}$ just because it works if $i$ is replaced by a real number. – Jason DeVito - on hiatus Feb 18 '23 at 03:19
  • We can start by defining $e^{-i\theta}$ as the complex-valued function satisfying the IVP $$f'(\theta)=-if(\theta),;f(0)=1.$$ For the sake of the proof, we only need to show $f(-\theta)f(\theta)=1$ (with Euler's formula, we can show $f(x+y)=f(x)f(y).)$ $$f(\theta)=\sum_{n=0}^\infty\frac{(-i\theta)^n}{n!}\ f(-\theta)=\sum_{n=0}^\infty\frac{(i\theta)^n}{n!}\ f(\theta)f(-\theta)=\sum_{n=0}^\infty \theta^n\sum_{k=0}^n\frac{(-i)^k}{k!}\frac{i^{n-k}}{(n-k)!}\ =\sum_{n=0}^\infty \frac{(i\theta)^n}{n!}\sum_{k=0}^n(-1)^k\binom nk\ =1+\sum_{n=1}^\infty\frac{(i\theta)^n}{n!}(1-1)^n\ =1.$$ – user7777777 Feb 18 '23 at 10:22
  • This might sound crazy, but it's what we have to do to make sure the pattern still holds in the complex realm, as pointed out by you, so it's actually not completely crazy if you think about it since sine and cosine solve the DE $$t''=-t$$which reduces to the characteristic equation $\lambda^2+1=0$ with ansatz $t=e^{\lambda\theta}.$ Anyway, now we can differentiate $f(\theta)$ in my answer where the complex derivative is the same as the real one for real-valued functions. – user7777777 Feb 18 '23 at 10:32
  • @FShrike I am assuming the relationship between the sides of the right triangle yes because I defined $\sin\theta$ and $\cos\theta$ as the legs of a right triangle with hypotenuse $1,$ such relationship has to exist because the lengths of two sides of a triangle define that of the third. What I am trying to do is find what the relationship between the sides is exactly. Never did I use the fact that the $x$ and $y$ coordinates of a point on the unit circle satisfy $x^2+y^2=1;$ I am trying to show it! – user7777777 Feb 18 '23 at 11:13
  • Yup, I see your point. I'm far too used to seeing the unit circle defined as $x^2+y^2=1$. I take back that last comment – FShrike Feb 18 '23 at 11:42
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Notice that in $a+bi=c e^{it}$, while it is straightforward to view $a,b$ as lengths of a right triangle, to claim the hypotenuse has length $c$ is exactly to claim that the complex modulus is the same thing as length,… which is circular, since that itself depends on the Pythagoras theorem (in other words, you are asserting $|e^{it}|=1$ hence the ‘length’ associated to $ce^{it}$ is just $c$).

FShrike
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I wanted to spell out my objection to the proof in a bit more detail.

For concreteness, when I'm thinking about a point in $\mathbb{R}^2$, I'll write it as $(x,y)$ as is customary. And to be clear, I'm thinking of $\mathbb{R}^2$ with its usual Euclidean vector space structure where we can measure lengths and angles as usual. I'm also viewing the distance function not as a formula, but as an input-output procedure. That is, I'm viewing it like a computer program where I type in $d((3,0), (0,4))$ and it spits out $5$. I don't get to see the inner workings of how it computes the answer.

On the other hand, we have this other algebraic structure $\mathbb{C}$. In the OPs proof, he/she attempts to prove something about $\mathbb{R}^2$ by proving something in $\mathbb{C}$ and then relating the two. To be clear, I'm thinking of $\mathbb{C}$ purely algebraically - there is no notion of distance or angles (yet).

It seems reasonable to assume that we want the identification of $\mathbb{C}$ with $\mathbb{R}^2$ to preserve some structure. Since both $\mathbb{R}^2$ and $\mathbb{C}$ are real $2$-dimensional vector spaces, it's reasonable to ask that any identification preserve this structure. Since $\{1,i\}$ forms a basis of $\mathbb{C}$ as a $\mathbb{R}$-vector space, the identification is determined once we identify $1$ and $i$ with points in $\mathbb{R}^2$.

Of course, the customary way is to identify $1$ with $(1,0)$ and $i$ with $(0,1)$. But there is nothing which forces us to adopt this.

Let's see what happens if we identify $1$ with $(1,0)$ and $i$ with $(1,1)$. Then a calculation reveals that that point $a+bi$ is identified with the ordered pair $(a+b,b)$. Going backwards, a point $(x,y)$ is identified with the complex number $(x-y)+yi$. Given $a\in\mathbb{C}$, I'll use $\vec{a}$ to denote its identified point in $\mathbb{R}^2$.

Notice that for the complex number $i$, if we write it in polar form as $1e^{i\pi/2}$, the number $1$ is not the length of $\vec{i}$. Indeed, $\vec{i} = (1,1)$ which has strictly longer length.

The obvious resolution is that whatever we identify $i$ with should have length $1$. So, instead of identifying $i$ with $(1,1)$, let's identify it with $\frac{1}{\sqrt{2}}(1,1)$. Then one can easily verify that $\vec{i}$ has length $1$.

Now, let's try to run the OP's proof in this modified setting.

Given a right triangle with legs of side lengths $4$ and $3$, we identify this (using my strange identification!) with the complex number $(4-3) + 3i = 1 + 3i$. Converting this to polar $1+3i = ce^{i\theta}$ and using the OPs line of computation, we conclude that $c = \sqrt{1^2 + 3^2} = \sqrt{10}$. This is true, but shows that the $c$ in the polar form is not the $c$ from the Pythagorean theorem.

The point is that we need something which tells us the $c$ in the polar form and the $c$ in the Pythagorean theorem are the same $c$.

Here is the something:

Fact: If we use the usual identification with $1 \mapsto (1,0)$ and $i\mapsto (0,1)$, then the two $c$s are the same.

So, if you know the Fact is true, then the OPs proof works to establish the Pythagorean theorem. Unfortunately, I do not know of a proof of the Fact which doesn't already use the Pythagorean theorem.

Jason DeVito - on hiatus
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  • Thank you! It’s really reassuring to see at least someone else agrees with my objections. +1 to this – FShrike Feb 18 '23 at 21:28
  • This is a very good way to demonstrate why implicit assumptions need to be reckoned with. – FShrike Feb 18 '23 at 21:30
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    @FShrike: I didn't read your answer carefully prior to mine. I now see we have the same objection. – Jason DeVito - on hiatus Feb 18 '23 at 21:38
  • The problem is fixed by Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta.$$ It infers that, using the definitions for sine and cosine, the real and imaginary axes are perpendicular with $(0,1)\equiv i$ and $(1,0)\equiv1.$ Euler's formula tells us how complex numbers sit on the complex plane. – user7777777 Feb 19 '23 at 08:25
  • Now, that we have shown that the real and imaginary parts of a complex number are "perpendicular vectors," we can proceed using the OP's proof by scaling both by a factor of $c,$ which enables us to make any right triangle for which we prove the pythagorean theorem $$a+bi:=c\cos\theta+ic\sin\theta\ =c(\cos\theta+i\sin\theta)\triangleq ce^{i\theta}\ (a+bi)(a-bi)=ce^{i\theta}\cdot ce^{-i\theta}\ a^2+b^2=c^2.$$ – user7777777 Feb 19 '23 at 08:25
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    @user7777777: I don't understand how Euler's formula implies that $1$ and $i$ are perpendicular. Euler's formula is an accurate statement about complex numbers regardless of how they're identified with $\mathbb{R}^2$. I agree that one definition of sine and cosine involves sides of a right triangle, but what does that imply a particular interpretation? Do you think a mass attached to an oscillating spring infers a right triangle because trig appears when modeling its motion? – Jason DeVito - on hiatus Feb 19 '23 at 12:44
  • @user7777777 (echoing Jason) I have the same objection with your most recent comment as I have had with most of the comments under my post. $a+bi=ce^{it}$ is true for some $t$, ok. If so, then it is also true that $a^2+b^2=c^2$. Ok... but what on Earth makes us link $c$ to the hypotenuse of the triangle? The pure mathematical algebra, calculus, series expansions, etc. are not a priori "geometric". You need to do the work to make that connection. The Argand diagram is a totally arbitrary fiction. I know I said I agreed with your posted answer, but I'm no longer sure. This is subtle. – FShrike Feb 19 '23 at 15:23
  • @FShrike I did not expect the proof to be right, I wanted to understand what is missing. You have multiple comments that are not clear to me and clutter the discussion. I would like to thank Jason DeVito for showing where it goes wrong. Is there a clear way to show that a point in the complex plane $a+bi$ is similar to a right triangle in the euclidean plane $\mathbb{R^2}$? – bananenheld Feb 19 '23 at 16:49
  • @bananenheld I'm sorry if my comments were unclear. From my viewpoint, I made a great effort to be precise about what the issue is. I only ever really made the same point, this can hardly be called "cluttering" – FShrike Feb 19 '23 at 16:58
  • @bananenheld: I don't know of a way of showing "a point in the complex plane $a+bi$ is similar to a right triangle" without already knowing the Pythagorean theorem. To emphasize it a bit, one should view my previous sentence as a statement about my lack of knowledge - I don't mean to imply that no one knows how to complete the task. But in order for your proof to be valid, this task must be completed without invoking the Pythagorean theorem. – Jason DeVito - on hiatus Feb 19 '23 at 18:04