TITLE: How fundamental is Euler's identity, really?
QUESTION [7 upvotes]: Euler's identity, obviously, states that $e^{i \pi} = -1$, deriving from the fact that $e^{ix} = \cos(x) + i \sin(x)$. The trouble I'm having is that that second equation seems to be more of a definition than a result, at least from what I've read. It happens to be convenient. Similarly, the exact nature of using radians as the "pure-number" input to trig functions is a similar question of convenience -- would it be fundamentally wrong to define sine and cosine as behaving the same way as they do now, except over a period of $1$ rather than $2 \pi$? In such a system, $e^{i \pi} = \cos(\pi) + i\sin(\pi) = \cos(\pi - 3) + i\sin(\pi - 3)$, or transforming back into our $2\pi$-period system to get a result, $\cos(\pi\frac{(\pi - 3)}{1}) + i\sin(\pi\frac{(\pi - 3)}{1})$, which is approximately $0.903 + 0.430i$. (Hopefully I did that right.)
Since there are equally mathematically true systems where $e^{i \pi}$ gives you inelegant results, I'm asking whether the fact that $e^{i \pi} = -1$ really demonstrates some hidden connection between $e$ and $\pi$ and the reals and imaginaries, as it rests largely on what seem to me to be arbitrary definitions of convenience rather than fundamental mathematical truths.

REPLY [6 votes]: One of the more remarkable things about this identity is that it falls out of so many different definitions of the terms. Its not just a convenience or a happenstance, it arises from almost any valid definition of exponentiation.
So, for instance, consider the definition $$e^x = \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n$$ which is historically where $e$ first arose, in the work of Jacob Bernoulli.
So now we can ask: does this definition lead to Euler's identity? Or, more explicitly, is $$\lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n = \cos(x) +i\sin(x)\ ?$$ Of course here, and later, we use the radian version of the trig functions, and $x\in \mathbb{R}$.
To answer this, lets assume that $|zw| = |z||w|$ and $\arg (zw) = \arg(z) +\arg(w) (\mod 2\pi)$. We can derive these identities using algebra, and results from geometry that are more than 2000 years old. Furthermore, these functions are continuous, which is obvious for $|\cdot|$ and is true for $\arg$ in the correct topology.
We can now calculate the modulus of the relevant limit.
$$\begin{align*} |e^{ix}| &= \left| \lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n \right|\\
&= \lim_{n\to\infty} \left| \left(1 + \frac{ix}{n}\right)^n \right| \\
&= \lim_{n\to\infty} \left| \left(1 + \frac{ix}{n}\right)\right|^n \\
&= \lim_{n\to\infty} \left( 1 + \frac{x^2}{n^2} \right)^{n/2} \\
&= \lim_{n\to\infty} \left(\left( 1 + \frac{x^2}{n^2} \right)^{n^2}\right)^{1/2n} \\
&= \lim_{n\to\infty} \left(e^{x^2}\right)^{1/2n} \\
&= 1
\end{align*}$$
The first line is our definition, the second is justified by continuity, the third by our modulus identity, the fourth by the definition of modulus, and from there we play with exponents and use our definition of the exponential (there's another way to do it with logs, but this ought to be fine).
We can also calculate the argument.
$$\begin{align*} \arg(e^{ix}) &= \arg\left( \lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n \right)\\ 
&= \lim_{n\to\infty} \arg\left(\left(1 + \frac{ix}{n}\right)^n \right)\\
&= \lim_{n\to\infty} n \arg\left(1 + \frac{ix}{n}\right)\\
&= \lim_{n\to\infty} n \arctan\left(\frac{x}{n}\right) \\
&= \lim_{h\to\infty} \frac{ \arctan(xh) - \arctan(0) }{ h } \\
&= \left. \frac{\text{d}\arctan'(xt)}{\text{d}t}\right\vert_{t=0} \\
&= x.
\end{align*}$$ the justifications here are much the same as before, with a little calculus thrown in at the end.
Taking our two results together, and using a little more geometry, we have that $$e^{ix} = \lim_{n\to\infty} \left(1 + \frac{ix}{n}\right)^n = \cos(x) + i \sin(x)$$ and by implication $$e^{i\pi} = \lim_{n\to\infty} \left(1 + \frac{i\pi}{n}\right)^n = -1$$. So, this isn't just some arbitrary thing, it appears with all the definitions of exponentiation that can be easily extended to the complex numbers.
Anyway, I hope this adds something to your understanding @Why-Seven-Six.