# Explain why $e^{i\pi} = -1$ to an $8^{th}$ grader?

I am an $8^{th}$ grader that is taking Algebra I. But nearly everyday I try to learn things outside of what I am learning in class.

Quite a while ago I discovered that $e^{i\pi} = -1$. This fascinated me although I do not understand it completely. I understand what imaginary and complex numbers are and what Pi is but do not completely understand Euler's number and why it has such an interesting property such as this.

Can someone explain this to me? (Remember, I'm in $8^{th}$ grade and may not understand completely. So if someone could 'dumb it down' for me that would be very appreciated.)

EDIT: It amazes me just how much people care on this site. Each and everyone of you are putting out amazing answers (I've read them all) but, having little knowledge of trigonometry and calculus, I feel like this question is just something that will have to wait.

Any other answers would be appreciated, but I feel as though I will not fully comprehend this until later years in school. Thank you, again, to everyone who made an effort to helping me understand :-).

• You have to first learn how to extend the function $e^x$ to the complex plane. There are several ways to do this but perhaps the most common is to use the power series. That may be a bit beyond an 8th grader though. – Gregory Grant Apr 20 '15 at 3:08
• why should an $8{th}$ grader know about $i,$ let alone $e^{i\pi}?$ – abel Apr 20 '15 at 3:08
• @abel As I said, I try to learn a lot of things outside of what I am learning currently in class. – Jamie Sanborn Apr 20 '15 at 3:09
• Starting a bit before Euler's time, mathematicians were investigating/discovering the relationship between the exponential function $\ e^x \$ and the trigonometric functions. What is known as Euler's Identity (although it was found around the time he was born) describes what happens when you allow the exponent to be an imaginary number: $\ e^{ix} \ = \ \cos x \ + \ i \ \sin x \$ . If you use $\ x \ = \ \pi \$ , you get the famous equation (some call it one of the most beautiful in mathematics). To understand this relation is a bit beyond Algebra I, however. – colormegone Apr 20 '15 at 3:09
• @Jamie: FWIW, I first saw this fact outside of class in about 7th grade, saw it "formally" for the first time in pre-calculus, got a first non-rigorous proof in Calculus II, and wasn't able to prove it "for real" until Analysis (in university). So while the question is a good one to be asking, you shouldn't be too discouraged if there is no overlap between answers you understand and answers you think are good. – Eric Stucky Apr 20 '15 at 3:20

Instead of proving it, I'll just explain how we can represent any complex number in polar form.

Any complex number can be represented in Cartesian form (the form you're probably familiar with) as $x+iy$, where $x,y \in \Bbb R$ (that means that $x$ and $y$ are real numbers) and $i^2 = -1$.

However all complex numbers can be represented in another form as well: polar form. Polar form involves specifying the distance from the origin and the angle that the ray through the point makes with the positive $x$ axis. We can represent complex numbers by $re^{i\theta}$ where $r$ is the distance, $\theta$ is the angle, and $i$ is the same as the one above.

To convert between the two, you use Euler's formula $$re^{i\theta}=r(\cos(\theta) +i\sin(\theta)) = x+iy$$

Here's an image that might help: Now you can see for yourself that $e^{i\pi} = 1e^{i\pi}$ has a length of $1$ and makes an angle of $\pi$. So it lies along the $x$-axis, $1$ unit to the left of the origin: i.e. it is $-1$.

• While it is still very hard to wrap my head around this, I think this answer is pushing me much further to understanding Euler's Identity. – Jamie Sanborn Apr 20 '15 at 3:23
• I'm surprised the OP accepted this answer, because it side-steps the main issue (though it is still a fine answer). To the point: the OP wrote that he "[does] not completely understand Euler's number and why it has such an interesting property such as this", to which you answered with Euler's formula, without giving any sort of explanation for it (intuitive, formal, or otherwise). – A.P. Apr 22 '15 at 21:55

If you graph $$y = e^x$$ and $$y = 1 + x,$$ the graphs look sort of similar near $x = 0$. If instead of $1 + x$ you graph $y = 1 + x + \frac{x}{2!}$, it looks even better. And if you graph $$y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^{10}}{10!},$$ it looks even better. If you think about a polynomial with infinitely many terms, following the same pattern, you can (with some real effort!) make sense of it, sort of the same way you make sense of $1 + (1/2) + (1/4) + \ldots = 2$. And that infinite polynomial will have a graph that looks exactly like the graph of $y = e^x$. So people say that $e^x$ and that infinite polynomial are "the same thing." (We call the polynomial a "power series" representation, because it involves successive powers of $x$, and it's an infinite series.)

Once you've said that $e^x$ and the power series are the same thing, it makes sense to talk about $e^z$, where $z$ is complex: you just use the polynomial to evaluate it.

You can do the same kind of thing for $\sin$, getting $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} \pm \ldots$$ and $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4}! \pm \ldots$$

[By the way... the way you guess these polynomials is to use calculus which isn't really what you were hoping for, I suppose. But among all first-degree polynomials, $y = 1 + x$ is the one that best fits $y = e^x$ near $x = 0$: the value at $0$ is correct, and the slope at zero is also correct. The higher-degree representations similarly are "best fits", but in more complicated ways.]

Once you've done that for sine and cosine, you can do some algebra to show that $$e^{ix} = \cos x + i \sin x$$ although you need to rearrange terms a lot, and proving that this is valid for these particular infinite polynomials is remarkably subtle.

But from that last great formula, you can plug in $x = \pi$ to get Euler's formula.

• does the op even know what a function is? – abel Apr 20 '15 at 3:26
• The OP is near the end of Algebra I, and should have heard of a "function" by now, yes. I confess, I haven't looked at Alg I curriculum in a while, but OP mentions using a calculator, thus knows about "function keys" at the very least. And if OP is curious enough to learn about Euler's formula, I'm pretty OP's encountered functions during previous investigations. – John Hughes Apr 20 '15 at 10:23
• @JohnHughes I'm learning new things everyday. I already do know what a function is. I also know a weird mix of things that an $8^{th}$ grader wouldn't although my knowledge overall is very foggy. For instance, I know how to integrate and derive polynomials, but I am clueless with just about everything else in (pre)-calculus. – Jamie Sanborn Apr 21 '15 at 19:52

As in my other answer, I won't be trying to prove anything to you -- power series, differential equations, etc. will be over your head for the moment. Instead I'll offer a different interpretation of the exponential function that hopefully you will find enlightening.

Instead of seeing $e^{i\theta}$ as a number -- let's interpret it as an operator on complex numbers. Think of it as a generalization of the negation operator "$-$". Geometrically, what negation does is it keeps the length of the arrow pointing from the origin to a number on the real number line the same, while flipping its direction. On the real number line there are only two different directions from the origin. So negation is a pretty easy thing to pin down. Not so with complex numbers. Graphing complex numbers requires us to plot on a plane. So from the origin there are an infinite number of different directions. So, not only could one potentially flip the direction of the arrow pointing to a complex number -- one could rotate it. That's what the exponential function $e^{i\theta}$ does -- it rotates a complex number about the origin by an angle $\theta$ (measured in radians, not degrees) without changing its length.

Let's look at an example. We'll start with the complex number $z=1 + i$. That one looks nice and easy. Now let's say that we want to rotate $z$ counterclockwise by an angle of $\pi /6$ to get a new complex number $z'$. Then we just use the exponential operator:

$$z' = e^{i\pi/6}z = e^{i\pi/6}(1+i)$$

Now actually finding an explicit form for $z'$ will require some geometry and trigonometry -- but it shouldn't be too hard. Let's give it a try.

First we'll graph $z$: From the Pythagorean theorem, we can see that the hypotenuse of that triangle is $\sqrt{1^1+1^2} = \sqrt{2}$. And from the fact that $$\cos(\theta) = \dfrac{\text{adjacent}}{\text{hypotenuse}}$$ we see that $\theta = \operatorname{arccos}\left(\frac 1{\sqrt{2}}\right) = \pi/4$.

Now we want to rotate that arrow (the hypotenuse of the triangle) that we're using to point to our complex number by $\pi/6$ while keeping its length the same. Well we can immediately see the angle that we'll get -- it'll just be $\pi/4 + \pi/6 = 5\pi/12$. So our new complex number $z'$ should have a picture that looks like this: Now using some trig, we see that the base of that triangle should be $\sqrt{2}\cos\left(\dfrac {5\pi}{12}\right) = -\dfrac 12 + \dfrac {\sqrt{3}}2$ and the height should be $\sqrt{2}\sin\left(\dfrac {5\pi}{12}\right) = \dfrac 12 + \dfrac {\sqrt{3}}2$.

Therefore $$z' = e^{i\pi/6}(1+i) = \left(-\dfrac 12 + \dfrac {\sqrt{3}}2\right) + \left(\dfrac 12 + \dfrac {\sqrt{3}}2\right)i$$

Now that we see how it works, let's consider the formula $$e^{i\pi} = -1$$ It looks like we can't use our approach -- afterall what is $e^{i\pi}$ operating on? What's it's operating on is the complex number $1=1+0i$. That is $e^{i\pi}$ is exactly the same as $e^{i\pi}(1)$.

So now it should be clear what we need to do -- we need to rotate the complex number $1$ by the angle $\pi$ radians. But $\pi$ radians is just halfway around a full circle. So $e^{i\pi}(1)=-1$. This may not be an answer suitable to what you've learned so far, but there's a nice way to relate the $e^{i\theta}$ to sines and cosines. When you get to calculus, you'll learn about the Taylor series representation for differentiable functions. It turns out that for any real number $x$ you can write $\sin x$ and $\cos x$ in terms of infinite sum of numbers. It turns out that

\begin{eqnarray*} \sin x & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \\ \cos x & = & 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} - \ldots \end{eqnarray*}

and similarly $e^x$ has an expansion

$$e^x \;\; =\;\; 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$$

If you add up $\cos x + i \sin x$ in terms of their series representation you get a nice result which is

\begin{eqnarray*} \cos x + i \sin x & = & 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots \\ & = & 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \ldots \\ & = & e^{ix} \end{eqnarray*}

thus giving us Euler's famous equation

$$e^{ix} \;\; =\;\; \cos x + i\sin x$$

I understand that the preceding may be a little much for someone to understand if you don't know what series are or where they come from, but once we have Euler's equation we have a nice interpretation: If you think of the complex plane as being numbers that lie on the real axis and the imaginary axis we have that $e^{ix}$ has real component $\cos x$ and imaginary component $\sin x$, and as $x$ increases (remember that $x$ is a real number) then we trace out the arc of a circle starting at the point $(1,0)$ (notice $e^0 = \cos 0 + i \sin 0 = 1$), and this goes counterclockwise in the positive imaginary part of the plane and when we go $\pi$ radians we reach the point $e^{i\pi} = \cos \pi + i \sin \pi = -1$.

Maybe this will work, without giving you all details, but, of course, feel free to ask any followup:

The identity $e^{i\pi}=-1$ denotes the fact that , in a(given) coordinate system, the Real number $-1$ forms an angle of $\pi$ radians (with the positive Real axis.) .The function $e^z$ takes in a Complex number in polar representation and returns the(a) number that has that polar representation. In this sytem, the Complex number whose polar rpresentation is $i\pi$ is the number $-1$. Expanding:

Polar numbers are often expressed either as an expression $x+iy$, with $x,y$ both Real numers, or in the form $r(cos\theta+isin\theta)$, where $\theta$ is the(an) angle formed by the Complex number with the axes. This angle is not well-defined unless you set some constraints on it. This last expression/representation is called the polar form of a Complex number.

The complex exponential $e^z=e^{x+iy}$ is a function that assigns to $z=re^{i\theta}$ (0ne of the) numbers with that polar representations. The polar representation of a number z consists of information on the length of $z$ (distance from the origin) as well as the angle formed with the origin. We consider the positive part of the Real axis to be the angle origin for numbers ( note that we can consider any line to be the angle origin, but our choice of the Real axis is convenient for some reasons). Then, since the positive Real axis is the angular origin and we consider angles counterclosckwise, i.e., the direction opposite that of the hands of a clock (for those that remember analog clocks) , the number $-1$ forms an angle of $\pi$ units from the origin on the Real axis, i.e., $e^{i\pi}=-1$ tells you that the Complex number ( in a specific coordinate system) that has length 1 and forms an angle of $\pi$ with the positive Real axis is the Real number (also Complex number) $-1$.

Euler's equation

$$e^{i\pi}+1 = 0$$

is considered by many to be the most beautiful equation in mathematics—rightly, in my opinion. However, despite what Gauss might say, it's not the most obvious thing in the world, so let's perhaps try to sneak up on it, rather than land right on it with a bang.

It's possible to think of complex numbers simply as combinations of real values and imaginary values (that is, square roots of negative numbers). However, plotting them on the complex plane, as Bye_World did, provides a kind of geometric intuition that can be valuable.

On the complex plane, a complex number $a+bi$ is plotted at the point $(a, b)$. Adding complex numbers is then just like adding vectors—$(a+bi) + (c+di) = (a+c)+(b+d)i$, for instance—just as you might have expected. (It's probably useful to draw some of these out on graph paper, if you can.)

Multiplication is where things get a little unusual. Multiplication by real values is just as you'd expect, generalizing from the one-dimensional real number line to the two-dimensional complex plane: Just as $k$ times a positive number is (for positive $k$) another positive number $k$ times as far from the origin, and correspondingly for negative numbers, $k$ times a complex number is another complex number, $k$ times as far from the origin, and in the same direction.

But multiplication by imaginary values is different. When you multiply something by $i$, you don't scale that something, you rotate it counter-clockwise, by $90$ degrees. Thus, the number $5$, which is $5$ steps to the east (so to speak) of the origin, when multiplied by $i$ becomes $5i$, which is $5$ steps to the north of the origin; and $3+4i$, which is to the northeast, becomes $-4+3i$, which is to the northwest. And so on.

OK, let's step away from the complex plane for a moment, and proceed to the exponential function. We're going to start with the ordinary ol' real-valued exponential function, $y = e^x$. There are lots of exponential functions: $2^x, 10^x, \pi^x, \ldots$ But there's something special about the exponential function with $e$, Euler's constant, as its base.

If you graph $y = e^x$, you get a curve that starts out at the far left, at $(-\infty, 0)$ (so to speak), and proceeds rightward, crawling very slowly upward, so slowly that by the time it gets to $x = 0$, it's gotten no further upward than $(0, 1)$. After that, however, it picks up speed, so that further points are $(1, e), (2, e^2), (3, e^3), \ldots$, and by the time $x = 20$, we've nearly halfway to a billion.

Another way to put that is that the derivative of $y = e^x$, which you might think of as its slope, starts out as an almost vanishingly small number far to the left of the origin, but becomes very large when we get to the right of the origin.

To be sure, all exponential functions do that basic thing. However, the very unusual thing about $y = e^x$ is that its derivative—its slope, in other words—is exactly itself. Other exponential functions have derivatives that are itself multiplied by some constant. But only the exponential function, with $e$ as its base, has a derivative that is exactly equal to itself.

It's very rare that an expression has that property. The function $y = x^2$, for instance, has derivative (or slope) $y' = 2x$, which is not equal to $x^2$. But if you want to know the slope of $y = e^x$ at any point, you just figure out what $y$ is, and there's your slope. At $x = 1$, for instance, $y = e \doteq 2.71828$, so the slope there is also $y' = e \doteq 2.71828$.

The only functions that have that property have the form $y = Ce^x$, where $C$ is any constant.

There's another way to think of the derivative that is not the slope, although it's related. It has to do with the effect that incremental changes in $x$ have on $y$. As we saw above, the derivative of $y = e^x$, at $x = 1$, is also $y' = e^x = e \doteq 2.71828$.

That means that if you make a small change in $x$, from $1$ to $1+0.001 = 1.001$, then $y$ approximately makes $2.71828$ times as much of a change, from $2.71828$ to $2.71828+0.00271828 \doteq 2.72100$. This is only accurate for small changes, the smaller the better, and in this case at least is exact only in the limit, as the change approaches zero. That is, in fact, the definition of the derivative.

Now, let's return to the complex plane, and put the whole thing together. Let's start with $e^0 = 1$. We can plot that point on the complex plane, and it will be at the point with coordinates $(1, 0)$. It's important to remember that this does not mean that $0 = e^1$. The value of $x$ is not being plotted here; all we're doing is plotting $y = e^0 = 1 = 1 + 0i$, and that $1$ and $0$ are the coordinates of $(1, 0)$, which is one step east of the origin. By the unusual property of $e^x$, the derivative is also $1$.

Suppose we then consider making a small change to $x = 0$. If we add $0.001$ to $x$, we make a change to $e^x$ that is equal to the derivative times the small change in $x$. That is to say, we add the derivative $1$ times the small change, $0.001$, or just $0.001$ again. So the new value would be close to (though not quite exactly) $1.001$, which is represented by the point $(1.001, 0)$. It would be in the same direction from the origin—east—as the original point, but $0.001$ further away.

But what happens if we add not $0.001$ to $x$, but $0.001i$? The derivative is still $1$, so the incremental impact on $e^x$ is the derivative $e^x = 1$ times $0.001i$, or $0.001i$ again. So the new value would be close to (though, again, not quite exactly) $1+0.001i$, which is represented by the point $(1, 0.001)$. It would be $0.001$ steps to the north of $(1, 0)$, because the extra factor of $i$ rotates the increment counter-clockwise by $90$ degrees.

Symbolically, we would say

$$e^{0.001i} \doteq 1+0.001i$$

Now, suppose we added another $0.001i$ to the exponent, so that we are now evaluating $e^{0.002i}$. We'll do what we did before, which was to multiply the increment in the exponent, $0.001i$, by the derivative. And what is the derivative? Is it $1$, as it was before? No, since we're making an incremental step from $e^{0.001i}$, it should be the derivative at $0.001i$, which is equal to $e^{0.001i}$ again, which we determined above to be about $1+0.001i$. If we multiply this new derivative value by the increment $0.001i$, we get an incremental impact on $e^x$ of $-0.000001+0.001i$, which is a tiny step that is mostly northward, but which is also just an almost infinitesimal bit to the west (that's the $-0.000001$ bit). We've veered ever so slightly to the left, so the new estimated value at $x = 0.002i$ is

$$e^{0.002i} \doteq 0.999999+0.002i$$

One thing to observe about the small steps that we've taken is that each one is at right angles to where we are from the origin. When we were directly east of the origin, our small step was directly northward. When we were just a tiny bit north of east from the origin, our small step was mostly northward, but a tiny bit westward, too.

What curve could we put around the origin, such that if we traced its path, the direction we're moving would always be at right angles to our direction from the origin? That curve is, as you might have guessed already, a circle. And since we start off $1$ step east of the origin, the circle has radius $1$. Unsurprisingly, this circle is called the unit circle.

If we follow this line of reasoning, then the value of $e^{i\pi}$ must be somewhere along this unit circle; that is, if $e^{i\pi} = m+ni$, then $m^2+n^2 = 1$ (since that's the equation of a circle of radius $1$, centered at the origin). The only reason our estimated values weren't exactly on the unit circle is that we made steps of positive size, whereas the derivative is technically good only for steps of infinitesimal size. But where on the unit circle is $e^{i\pi}$?

The crucial observation is in how fast we make our way around the circle. When we made our first step, from $x = 0$ to $0.001i$, that step had a size, a magnitude, of $0.001$, and the incremental impact on $e^x$ was also of magnitude $0.001$. Our second step, from $x = 0.001i$ to $0.002i$, was also of magnitude $0.001$, and the incremental impact on $e^x$ was, again, about $0.001$.

In order to get to $e^{i\pi}$, we would have to make a bunch of steps, whose combined magnitude total $\pi$. The result would be, if we reason as we did above, to move a distance $\pi$ around the unit circle. Since the unit circle has radius $1$, and diameter $2$, its circumference must be $2\pi$. Therefore, $e^{i\pi}$ must be halfway around the circle, at coordinates $(-1, 0)$. That is none other than the complex value $-1+0i = -1$:

$$e^{i\pi} = -1$$

or, in its more common form,

$$e^{i\pi}+1 = 0$$

The foregoing is not, by any means, a rigorous demonstration. It's an attempt to give some kind of intuition behind the mysterious-looking formula. I hope that helps.

P.S. I see you've already accepted an answer. Ahh well. I hope it helps, anyway.

• Bye_World's answer, in my opinion, was easier to comprehend. But when you were talking about taking 'pi' steps to the east to approach -1, this made me picture this whole thing a lot clearer and I thank you. :-) – Jamie Sanborn Apr 20 '15 at 4:50
• Bye_World was, perhaps, a little less ambitious. :-) As he/she said, the answer was not so much to demonstrate, as to help visualize. It's good to get multiple perspectives, on the notion that one understands things in more than one way. And at any rate, there may be others who don't really understand it, either. – Brian Tung Apr 20 '15 at 4:59

[I see that Brian Tung has already posted a similar answer to this one. Although this is shorter, it's saying essentially the same thing, so I don't know if it will be easier to understand. Perhaps harder!]

If:

(1) you're familiar with the operation of differentiation of real-valued functions of real numbers;

(2) you know that the function $x \mapsto e^x$ is the only function that satisfies the differential equation $f'(x) = f(x),$ subject to the initial condition $f(0) = 1;$

(3) you know the Chain Rule, at least well enough to know its consequence that $\frac{d}{dx}e^{ax} = ae^{ax},$ for any real number $a;$

(4) you can take it on trust (or already know!) that differentiation of complex-valued functions of a real variable, and even (under certain conditions) differentiation of complex-valued functions of a complex variable, can be defined in much the same way as for real-valued functions;

(5) you can take it on trust that the function $x \mapsto e^x$ can be extended to a complex-valued function of a complex variable, $z \mapsto e^z,$ and that it still satisfies the same differential equation;

(6) you can believe that the Chain Rule still applies, so that it follows that $\frac{d}{dt}e^{it} = ie^{it},$ for all real $t;$

(7) you're familiar with the representation of complex numbers as points of the Euclidean plane;

(8) in particular, you know that multiplying a complex number by $i$ corresponds geometrically to rotating it through a quarter-turn anticlockwise;

(9) you can see from Euclidean geometry that, if $\mathbb{U}$ is the circle of unit radius centred on the origin $0 = (0, 0),$ and if $f(t)$ denotes the point of $\mathbb{U}$ at arc length $t$ anticlockwise from the point $1 = (1, 0),$ then clearly $f(0) = 1,$ and (intuitively, by looking at a diagram for a "small" increment in the value of $t$) $f'(t) = if(t),$ for all real $t;$

then you will at least find it plausible that $f(t) = e^{it}$ for all real $t,$ and in particular: $$-1 = (-1, 0) = f(\pi) = e^{i\pi}.$$

That's a lot of "if"s for an 8th-grader. (But this is how I "dumb it down" so I can understand it!)

Step (9) is perhaps too condensed. But the idea is simply that a point moving around a circle has an instantaneous "velocity", with respect to arc length, which is a dimensionless unit vector pointing along the tangent to the circle, in the direction of the motion, at right angles to the current radius.

At the time of my writing, there are 7 lengthy, beautifully written answers. They all, in one way or another, take as a launching point the definition of $e^x$ as an infinite series, or as a representation of a complex number in polar form on the complex plane. What an eight-grader might find puzzling is why one would approach e in this manner.

There is a definition that has yet to be touched upon. It ties together the two streams, infinite series and geometrical representation. It is in fact the original definition, the way it was actually discovered.

To start at the very beginning with Napier's work on the logarithm would take us a little too far afield, so let's begin a bit later with Jacob Bernoulli. As a mathematician who was always short on cash, he was quite interested in the mathematics of finance.

I'm sure this has been gone into at length in other posts on Math StackExchange, so I'll be brief. Basically the idea is that compounding interest $n$ times on \$1 leads to the formula$(1 + r/n) ^n$. This formula represents paying$100*r $% interest on \$1. So if you pay 100% interest ($r = 1$) then $(1 + 1/n)^n$ is the total amount, interest and principal, you owe at the end of the $n$-th interest period.

Bernoulli wanted to know what happened when $n$ was infinity, i.e. you were continuously compounding interest. The answer is that as $n$ approaches infinity, the above expression approaches $e^r$. It's possible, but not recommended, to use this to compute $e$.

This is how $e$ was discovered. Because someone wanted to know if there was a ceiling on the amount of interest you could earn if you increased the number of compounding periods.

Now what about $e^{i\pi}$? Given that $e^r = lim_{n\rightarrow \infty} (1+r/n)^n$, the natural thing to do is plug in $i\pi$ for $r$. This means you're looking at the limit of $(1 + i\pi/n)^n$. Now there are two things you can do. You can use the binomial theorem, which tells you how to expand algebraic expressions like this, or you can use complex geometry. I will do the latter. If you use the former, you're going to have to use infinite series for cosine and sine at some point.

From the point of geometry though, the answer is obvious. $1 + i\pi/n$ is very close to 1 on the complex plane. In fact, it is on the tangent line to the point (1,0) on the unit circle. This means that for $n$ large, the point is very close to the point on the unit circle that is $\pi/n$ radians counter-clockwise from (1,0). Call that point $u$. If you know that multiplying complex numbers is the same as adding their angles and multiplying their magnitudes, then you should see $u^n = -1$. Since the result of multiplication won't change much if the inputs are perturbed, $(1 + i\pi/n)^n$ is very close to -1 for large $n$, and will approach -1 as n goes to infinity. Therefore $e^{i\pi} = -1.$

Credit to Timothy Gowers and his Princeton Companion to Mathematics for emphasizing the compound interest approach. A good book for any mathematician, budding or long-in-the-tooth, to have on the bookshelf.

• This is my preferred argument. However, I once attempted to charm a second-term Calculus class in a liberal arts college with it, and the response was to ask me whether it was good for anything. If I had had my wits about me and the forethought, I would have prepared myself by being able to quote the Millay sonnet, “Euclid alone has looked on Beauty bare.” – Lubin May 26 '15 at 2:18
• We live in a strange world where engineering majors love complex exponentials but liberal arts majors ask "what good is it?" – Chan-Ho Suh May 26 '15 at 2:21