What follows is a bit I did over at Math StackExchange. Posting it over here was an experiment in whether the mathematical typesetting would transfer correctly in a copy-and-paste. For the most part, as long as I leave it alone, it seems to have done so (modulo the line breaks being lost in the shuffle).
Euler's equation
eiπ+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 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=ex. There are lots of exponential functions:
2x,10x,πx,… But there's something special about the exponential function with
e, Euler's constant, as its base.
If you graph
y=ex, you get a curve that starts out at the far left, at
(−∞,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,e2),(3,e3),…, and by the time
x=20, we've nearly halfway to a billion.
Another way to put that is that the
derivative of
y=ex,
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=ex 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=x2, for instance, has derivative (or slope)
y′=2x, which is not equal to
x2. But if you want to know the slope of
y=ex at any point, you just figure out what
y is, and there's your slope. At
x=1, for instance,
y=e≐2.71828, so the slope there is also
y′=e≐2.71828.
The only functions that have that property have the form
y=Cex, 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=ex, at
x=1, is also
y′=ex=e≐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≐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
e0=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=e1. The value of
x is not being plotted here; all we're doing is plotting
y=e0=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
ex, 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
ex 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
ex is the derivative
ex=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
e0.001i≐1+0.001i
Now, suppose we added another
0.001i to the exponent, so that we are now evaluating
e0.002i. 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
e0.001i, it should be the derivative at
0.001i, which is equal to
e0.001i 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
ex 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
e0.002i≐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
eiπ must be somewhere along this unit circle; that is, if
eiπ=m+ni, then
m2+n2=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 eiπ?
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
ex 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
ex was, again, about
0.001.
In order to get to
eiπ, we would have to make a bunch of steps, whose combined magnitude total
π. The result would be, if we reason as we did above, to move a distance
π around the unit circle. Since the unit circle has radius
1, and diameter
2, its circumference must be
2π. Therefore,
eiπ must be halfway around the circle, at coordinates
(−1,0). That is none other than the complex value
−1+0i=−1:
eiπ=−1
or, in its more common form,
eiπ+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.
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