Has anyone talked themselves into understanding Euler's identity a bit?

Question

What does the ratio of the circumference of a circle to its diameter have to do with the base of the natural logarithm and $\sqrt{-1}$?

Answer 1

The natural exponential function $w=e^z$ changes at a rate equal to its own value. So at $z=0$, we have $w=1$ and the rate of change $dw/dz$ is $1$. So (like Euler) consider an infinitely small but nonzero change in $z$, called $dz$. What is $dw=e^{0+dz}$? Since the rate of change is $1$, we have $dw=dz$. In particular that means that if $dz = i\;dy$ where $dy$ is infinitely small and real, then it's moving from $1$ in the complex plane in a direction that is straight up or straight down, i.e. along the unit circle!.

Now suppose $e^a$ is some point on the circle and think about $e^{a + i\;dy}$ where $dy$ is infinitely small and real. Again, we want the rate of change to be equal to the value of the function, and that means the rate of change is that same point on the circle. Draw a vector from $0$ to that point; then $w$ is changing just that many times as fast as $z$ is changing. Now notice that the infinitely small quantity $i\;dy$ is a pure imaginary, and when you multiply by a pure imaginary you rotate $90^\circ$. That means $dw$ will be just as big in absolute value as $dy$ (since the absolute value of the derivative at that point is $1$, since the point is on the unit circle) but rotated $90^\circ$. That means instead of an arrow from $0$ out to a point on the circle, hitting the circle at a right angle, we've got an arrow tangent to the circle. In other words, as $z$ changes in the real direction, $w$ is changing in a direction tangent to the circle at the point where $w$ is located.

So as $y$ moves along the real line and $iy$ along the imaginary axis, $w$ moves alonng the circle! (And at the same speed.)

Answer 2

Here's a viewpoint from the perspective: "Series made me do it."

If you take the definition of $e^x$ to be $\sum\limits_{n=0}^\infty \frac{x^n}{n!}$. Then $e^{ix} = \sum\limits_{n=0}^\infty \frac{(ix)^n}{n!}$. Keeping in mind $i^2=-1$, $i^3=-i$, etc., $e^{ix} = \sum\limits_{n\;\mathrm{even}} \frac{(ix)^n}{n!} + \sum\limits_{n\;\mathrm{odd}} \frac{(ix)^n}{n!} = \sum\limits_{k=0}^\infty \frac{(ix)^{2k}}{(2k)!} + \sum\limits_{\ell=0}^\infty \frac{(ix)^{2\ell+1}}{(2\ell+1)!}$ Thus $e^{ix} = \sum\limits_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k)!}+\sum\limits_{\ell=0}^{\infty} \frac{(-1)^\ell i x^{2\ell+1}}{(2\ell+1)!} = \cos(x)+i\sin(x)$ (recognizing the MacLaurin series of sine and cosine).

Therefore, $e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+i0=-1$.

So Euler says, "I couldn't help myself. Taylor series made me do it that way."

Answer 3

Let's try another approach, using infinitely small nonzero numbers and infinitely large integers, as Euler often did. Let $n$ be an infinitely large integer. Then $$ e^{iy} = \left(e^{iy/n}\right)^n = \left(1 + \frac{iy}{n}\right)^n. $$ Here's where the last identity came from: $w=e^z$ so $dw=e^z\;dz$, where $dz$ is an infinitely small change in $z$. In the case we're considering, $z$ changes from $0$ to $iy/n$, and at $z=0$ we have $e^z=1$ so $dw=dz= iy/n$. Now do a binomial expansion and wherever you see a term like $$ \binom{n}{6}\left(\frac{iy}{n}\right)^6, $$ you say $$ \begin{align} \binom{n}{6} & = \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!} \left(\frac{iy}{n}\right)^6 \\ \\ \\ & = \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{n^6} \left(\frac{iy}{6!}\right)^6 \\ \\ \\ & = \left(\frac{iy}{6!}\right)^6, \end{align} $$ where the last equality follows from the fact that $n$ is infinitely large (in modern parlance, the limit of that fraction as $n\to\infty$ is 1).

Then just as in Bill Cook's answer (which I upvoted), we split the real part from the imaginary part and recognize the series for sine and cosine.

Answer 4

How about some geometry:

Reals: $\mathbb{R}$ encode 1 dimensional geometry. Complexes: $\mathbb{C}$ encode 2 dimensional geometry. Quaternions: $\mathbb{H}$ encode 3 dimensional geometry.

"Addition = Translation" and "Multiplication = Scaling/Rotation"

Specifically, multiplication by a "unit length" number will perform a rotation. $e^{i\theta}$ rotates numbers by $\theta$ radians. Thus $e^{i\pi} \cdot 1 = -1$ (in the complex plane).

[Disclaimer: I don't pretend that this is a good way to arrive at this formula, but I do think it helps one understand how/why the formula "works".]