Is this a valid proof of Pythagoras theorem using complex numbers? 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?

 A: 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$}$
A: 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 $c$).
A: 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.
