Physical Proof of Euler's Formula

Question

I would like to construct a geometrical or physical proof of Euler's Formula $e^{ix}=\cos x +i\sin x $. If anyone has constructed such a proof before I would love to see it, if not, I would like some guidance. Currently I'm thinking of how Euler's formula could be arrived mapping imaginary numbers onto the unit circle, but I feel as thought something is missing; any help would be greatly appreciated.

Answer 1

If we want to define a function $f \colon \mathbb{R} \to \mathbb{C}$, $f(t) = e^{it}$ in a way that meshes with our formulas for the real exponential function, it makes sense to require that $f(0) = 1$ and $f'(t) = if(t)$. Let's prove that we must then have $f(t) = \cos t + i\sin t$.

Write $f(t) = x(t) + iy(t)$. We have $f'(t) = x'(t) + iy'(t)$. First note that $$(d/dt)|f(t)|^2 = (d/dt)(x(t)^2 + y(t)^2) = 2x(t)x'(t) + 2y(t)y'(t) = \operatorname{Re}\left[2\overline{f(t)}f'(t)\right] = \operatorname{Re}\left[2\overline{f(t)}if(t)\right] = 0$$ since $\overline{f(t)}f(t)$ is real. It follows that $|f(t)|$ is constant. Considering the initial condition $f(0) = 1$, this implies that $|f(t)| = 1$, so $f(t)$ moves around the unit circle.

Noting further that $|f'(t)| = |if(t)| = |f(t)| = 1$, this implies that $f(t)$ moves around the unit circle at a constant speed of $1$. Finally, since $f'(t) = if(t)$, it always moves at a velocity that is rotated $90^{\circ}$ counterclockwise from its radius vector.

In conclusion, $f(t)$ starts at $1$ and moves counterclockwise around the unit circle at a velocity of $1$. In other words, $f(t) = \cos t + i\sin t$ (as the stated conditions can practically be taken as a geometric definition of the sine and cosine functions).