Physical Proof of Euler's Formula

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.

• There's a geometric "proof" of Euler's formula here: en.wikipedia.org/wiki/Euler%27s_formula#Proofs – David G. Stork Jan 25 '15 at 5:58
• You should try Tristan Needham's Visual Complex Analysis. – hjhjhj57 Jan 25 '15 at 6:04
• From a logical standpoint, this is mostly a matter of definitions. So the question is, why is it natural to define $e^{ix}$ in such a way that this happens. Only after you've given the definition would you try to prove the formula. I think that if you're looking for a way to define function $f(t) = e^{it}$ from $\mathbb{R}$ to $\mathbb{C}$, it should have the properties that $f(0) = 1$ and $f'(t) = if(t)$. It then becomes a matter of solving a differential equation to show that $f(t) = \cos t + i\sin t$ is the only possibility. – user208259 Jan 25 '15 at 6:04
• I downloaded Needham's Visual Complex Analysis, is there one particular section I should look at or is the whole book something I should use to educate myself on this topic? – user1505399 Feb 3 '15 at 2:29

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).