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

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

-
The answers on this MO question are relevant. –  Zev Chonoles Oct 12 '11 at 19:57
sqrt(-1)=i (from complex number definition (a+i*b) –  dato datuashvili Oct 12 '11 at 19:59
I don't think it has so much to do with "the base of the natural logarithm" as it has with "the natural exponential function". –  Henning Makholm Oct 12 '11 at 20:37
I'm sure this is a duplicate... though my search skills are on the fritz today. –  Guess who it is. Oct 13 '11 at 1:09
If you plot the function $f(x) = e^{ix} = \cos x + i\sin x$ in the complex plane, then every point is of the form $(\cos x, \sin x)$, so the graph of the function is just the unit circle. Then, as you increase $x$ from $0$ to $2 \pi$ you're traversing the unit circle. The statement $e^{i \pi} = -1$ is then just a subtle way of saying that if you start at $(1, 0)$ (i.e. starting at $x = 0$) then if you travel $\pi$ radians around the unit circle, you will end up at $(-1, 0)$. Of course, this doesn't explain why there is a connection between $e^{ix}$ and the unit circle... –  nsanger Mar 1 '14 at 4:15

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.)

-
$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}e^{it}=ie^{it}$! The direction of movement is perpendicular ($i$) to the position, and the speed ($\left|ie^{it}\right|$) is $1$. This is very intuitive! –  robjohn Oct 13 '11 at 2:30
robjohn, I do not know what you mean to be perpendicular to the position. Do you mean that the direction (vector) is perpendicular to the position vector? –  Darrin Mar 1 '14 at 4:26
@Darrin What he is getting at is: if you multiply a complex point by $i$, $M = iN$, then the location of $M$ on the complex plane is $N$ rotated a quarter revolution counter-clockwise. So if point $N(t)$ is moving in the direction of $M(t)$, then it is always moving counter clockwise in a circle. –  DanielV Mar 1 '14 at 7:57

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."

-

One of the many equivalent characterizations / definitions of $e$ is this: $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ Therefore, to find $e^{i\pi}$, we can look at the quantity $$\left(1+\frac{i\pi}{n}\right)^n$$ for higher and higher values of $n$ and see what it approaches. Remember that multiplication of complex numbers works by adding angles and multiplying lengths, and observe that for any given $n$, we have $$\left\lvert1+\frac{i\pi}{n}\right\rvert>1,\qquad \mathrm{arg}\left(1+\frac{i\pi}{n}\right)>0$$ so that as the value of $k$ increases, the complex number $\bigl(1+\frac{i\pi}{n}\bigr)^k$ will rotate counterclockwise in the plane by equal steps while growing in magnitude geometrically, forming a spiral shape. For any given $n$, the value of $\bigl(1+\frac{i\pi}{n}\bigr)^k$ at $k=n$ is what we're interested in, as mentioned before.

(This animation was inspired by the animation from the Wikipedia page on Euler's identity. While mathematically great, I have long been unsatisfied with its quality - it is low resolution, not self-contained, and uses the default style of Mathematica. This is my endeavour to make a better version.)

My Mathematica code:

MyLabel[text_, location_] :=
FontFamily -> "Linux Libertine", 17, Bold], location]];

EulerPlot[n_] :=
Show[ListLinePlot[
Table[{Re[(1 + I Pi/n)^k], Im[(1 + I Pi/n)^k]}, {k, 0, n}],
PlotRange -> {{-2.6, 1.2}, {-0.2, 3.6}},
AspectRatio -> 1,
ImageSize -> {500, 500},
PlotStyle -> Red,
PlotMarkers -> Automatic,
AxesLabel -> {"Real part", "Imaginary part"},
AxesStyle ->
Directive[FontFamily -> "Linux Libertine", FontSize -> 15]],
MyLabel["Plot of \
\!$$\*FormBox[\(TraditionalForm\\*SuperscriptBox[\((1 + \ \*FractionBox[\(i \[DoubledPi]$$, $$n$$])\), $$k$$]\),
MyLabel["for \!$$TraditionalForm\k$$ from 0 to \
\!$$TraditionalForm\n$$", {-1.6, 3.02}],
MyLabel[DisplayForm[
RowBox[{"\!$$TraditionalForm\n$$", "\!$$TraditionalForm\ =$$ ",
n}]], {-1.6, 2.65}]];

Export["animation.gif",
Table[EulerPlot[n], {n, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30,
40, 50, 100}}], "DisplayDurations" -> {0.75}, ImageResolution -> 80]
`
-

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.

-

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".]

-

To me, the simplest way to see this is through the roots of unity. Yes, you should go through the proof that $e^{i\theta}=cos\theta+isin\theta$ with the Taylor series, but even after doing so, it may not be apparent why Euler's identity is true. To help visualize it, we can turn to the roots of unity, using the definition of $e^{i\theta}$ in terms of sine and cosine.

Let's define the $n^{th}$ root of unity as a solution of $x^n=1$. So $e^{i\theta}$ is a solution to this if and only if and only if $n\theta$ is a multiple of $\pi$, since we want cosine to be $\pm1$ and sine to be $0$. So the set of all solutions of $x^n=1$ is $e^{\frac{k 2 \pi i}{n}}$, with k running from $1$ to $n$.

Looking at the geometry of it, we can also see that the roots of unity correspond to points of the unit circle on the complex plane:

The x-axis corresponds to the real part, and the y to the imaginary. We can see what $e^{pi i}$ is geometrically by looking at this.

So if we let $n=2$, we see that $e^{\pi i}$ is a root, and that it corresponds to a point on the unit circle at $\theta=\pi$, which is just $(-1,0)$. So that is -1 on the real axis and 0 on the imaginary axis.

So in short, $e^{\pi i}=-1$ because it corresponds to (-1,0) on the unit circle in the complex plane, leaving no imaginary part (since $isin\pi=0$).

Beautiful stuff.

-