How fundamental is Euler's identity, really?

Solution 1:

The real function $\exp: \mathbb R \to \mathbb R$ is the unique solution to the initial value problem $f'(x)=f(x)$ and $f(0)=1$. The complex function $\exp: \mathbb C \to \mathbb C$ is defined exactly the same way - as the unique complex-differentiable function satisfying the same initial value problem. The fact that such a solution exists over $\mathbb C$ is not obvious at all; it has to be proven. The easiest way to prove it is using the Taylor Series expansion of $\exp(z)$. In fact this is the unique way of extending $\exp$ from $\mathbb R$ to $\mathbb C$ while keeping the condition that it be differentiable.

It just so happens that, when $\exp$ is defined this way, we get the identity $\exp(ix) = \cos(x)+i\sin(x)$, where $\cos$ and $\sin$ use radians. There is no choice here; this is simply what comes out of defining $\exp$ in the only manner possible suited for analysis.

To see this using Taylor series, you need to know what the Taylor series for $\exp$, $\cos$, and $\sin$ are, where $\cos$ and $\sin$ use radians. I'll assume these are known, and they are:

$$\exp(z) = \sum_{k=0}^\infty \frac{z^k}{k!} = 1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!} + \cdots,$$

$$\sin(z) = \sum_{k=0}^\infty (-1)^k\frac{z^{2k+1}}{(2k+1)!} = z - \frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\cdots,$$

and

$$\cos(z) = \sum_{k=0}^\infty (-1)^k\frac{z^{2k}}{(2k)!} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots.$$

The idea is simply to put $ix$ in for $\exp(z)$. I'll just show what you get with the first few terms:

$$\exp(ix) = 1+(ix)+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}+\cdots$$ $$=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots$$ $$=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)$$

where I've collected real and imaginary parts in the last step. The observation is simply that the real part matches the taylor series for $\cos(x)$ and the imaginary part matches the taylor series for $\sin(x)$.

Solution 2:

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.