Are there any calculus/complex numbers/etc proofs of the pythagorean theorem? I have been looking for proofs for the pythagorean theorem that don't use area calculation but calculus, complex numbers or any other interesting ways to proof it.
I would love to see any interesting proof,
Shay
 A: There is a proof using Similar Triangles:

We get from $\triangle CDA$ and $\triangle ABC$ that $\displaystyle \frac{CD}{AC} = \frac{CA}{BC}$ i.e. $\displaystyle \frac{\alpha}{a} = \frac{a}{c}$ i.e. $ \alpha c = a^2$
Similarly $\displaystyle \beta c = b^2$.
Adding gives the result.
A: Explanation in terms of linear algebra. From  this blog post by Terence Tao

The statement $a^{2}+b^{2}=c^{2}$ is equivalent to the assertion that the
  matrices $%
\begin{pmatrix}
a & b \\ 
-b & a%
\end{pmatrix}%
$ and $%
\begin{pmatrix}
c & 0 \\ 
0 & c%
\end{pmatrix}%
$ have the same determinant. But it is easy to see geometrically that the
  linear transformations associated to these matrices differ by a rotation,
  and the claim follows.

A: Isn't the answer is just what's in here: Proof using differentials ?
A: How about the simplest ever (I got it from a book): Imagine a right prism with the base the triangle in question A(right angle), B , C (counter clock wise). The height of the prism is arbitrary. Now fill the prism with a gas at a given pressure. On the faces, in their middle the pressure forces act (Surface * pressure, perpendicular on the face). Equate the momenta around the corner B. They should cancel (the prism does not rotate by itself). You get the Pitagora's theorem in a blink of an eye. Cheers!! PS: I wanted to upload the figure, but I am not allowed until I get more points!
A: http://www.cut-the-knot.org/pythagoras/CalculusProofCorrectedVersion.shtml
This proof also appears in The American Mathematical Monthly, April 2011 issue.
A: An elegant proof.  Consider the vector $\mathbf{c}$ in the complex plane: $\mathbf{c}= a + \mathbf{i}b$, where $a$ and $b$ are scalar values, and $a,b > 0$.  Now consider the reflection of $\mathbf{c}$ in the line at an angle of 45 degrees (pi/4 radians) passing through the origin, call that $\mathbf{d}$. $\mathbf{d} = b + \mathbf{i}a$, and $\mathbf{d}$ clearly has the same length as $\mathbf{c}$.
Now take the vector product of c and d.  That has the same length as the square of the length of c, and in terms of polar coordinates, it is at an angle of pi radians (by virtue of the multiplicative properties of complex numbers expressed as polar coordinates).
so $\mathbf{c\cdot d} = i(c^2) = (a+ib)\cdot (b+ia) = ab-ba+i(a^2 + b^2) = i(a^2 + b^2)$.
$c^2 = a^2 + b^2$
