Is this a sound demonstration of Euler's identity?

Solution 1:

So, just rearranging a bit the proof, the two functions: $$ f(z) = \cos(z)+i\sin(z),\qquad g(z)=e^{iz} $$ are indeed the same function since they both are solutions to the first-order ODE $$ h'(z)-i\cdot h(z) = 0 $$ with the initial condition $h(0)=1$. We are just exploiting the Cauchy-Lipschitz theorem.

By evaluating $f$ and $g$ at $z=\pi$ we get Euler's celebrated identity.

The proof is perfectly fine - but it is not really different from proving $f\equiv g$ by comparing their Taylor series, if we consider that Frobenius' power-series method is the most natural way for solving some ODEs. The same situation happens in a discrete setting, for instance. We may prove that:

$$ F_n = \sum_{0\leq k \leq \frac{n-1}{2}}\binom{n-1-k}{k} $$ holds by exploiting the binomial theorem, or by proving that both the LHS and the RHS satisfy the same recurrence relation with the same initial conditions.

Solution 2:

This is not a proof of Euler's discovery, but of something much weaker.

What Euler noticed is that there is a connection between the exponential function and the trigonometric functions. Your proof only shows that solutions to $F'(t)=iF(t)$ (for real $t$) are equivalent to uniform circular motion in the complex plane. This suggests some formal relation of $F(t)$ to exponentials, but it does not:

• give a construction of $F(t)$ independent of circular motion;
• show that $F(ix)$ exists for real $x$
• show that $F(-ix)$, for real $x$, equals the function we know as $e^x$.

Those (or their equivalents) are the earth-shaking facts that Euler uncovered and the reason his formula is celebrated.

Starting from a more advanced formalism (such as power series) where $e^z$ is defined for both real and imaginary $z$ then an argument like this one might reproduce Euler's finding.