Most complicated proof of Pythagoras

Question

Usually a mathematician aims for clarity and elegance when conducting a proof. However, the antimathematician buries all hope of assimilating intuition and reasoning. To illustrate this, I seek the most complicated and/or long-winded proof of the beloved Pythagorean theorem. It doesmt have to be War and Peace or Paradise Lost, but length is always appreciated.

Answer 1

I love the challenge.

1. $e^z=\sum_{n\geq 0}\frac{z^n}{n!}$ defines an entire function over the complex plane.
   By convolutions we have $e^z\cdot e^{w}=e^{z+w}$ for any $z,w\in\mathbb{C}$;

2. The elementary trigonometric functions can be defined, for any $\theta\in\mathbb{R}$, as $\sin\theta=\text{Im } e^{i\theta}$ and $\cos\theta=\text{Re }e^{i\theta}$. This is equivalent to the standard definition since the map $\theta\mapsto \cos\theta+i\sin\theta = e^{i\theta}$ gives a parametrization of $S^1$ with constant speed: the speed is constant since $\frac{d}{dz}e^z=e^z$ is a trivial consequence of termwise differentiation of the series defining $e^z$;

3. It is a parametrization of $S^1$ since for any $\theta\in\mathbb{R}$ we have $$ \|e^{i\theta}\|^2 = e^{i\theta}\cdot \overline{e^{i\theta}} = e^{i\theta}\cdot e^{\overline{i\theta}} = e^{i\theta}\cdot e^{-i\theta} = e^0 = 1.$$

4. In terms of $\sin\theta$ and $\cos\theta$, the last identity takes the form $$(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)=\color{red}{\cos^2\theta+\sin^2\theta = 1}$$ which is the Pythagorean theorem.