Why does $e^{i\pi}=-1$?

Question

I will first say that I fully understand how to prove this equation from the use of power series, what I am interested in though is why $e$ and $\pi$ should be linked like they are.

As far as I know $\pi$ comes from geometry (although it does have an equivalent analytical definition), and $e$ comes from calculus.

I cannot see any reason why they should be linked and the proof doesn't really give any insights as to why the equation works.

Is there some nice way of explaining this?

The animated image to the right hand side of this article, along with the explanation below, might also be of some assistance. — Lucian, Jun 08 '14 at 14:48
I suggest you read some complex analysis book. And I think you will find the answer. It's all about the definition which gives this. And if you go deeper you will find the definition is a natural way. — Schrödinger's Cat, Jun 08 '14 at 14:00

score 0 · Accepted Answer · answered Jun 15 '14 at 19:58

Euler's formula describes two equivalent ways to move in a circle.

Starting at the number $1$, see multiplication as a transformation that changes the number $1 \cdot e^{i\pi}$.
Regular exponential growth continuously increases $1$ by some rate; imaginary exponential growth continuously rotates a number in the complex plane.
Growing for $\pi$ units of time means going $\pi\,\rm radians$ around a circle
Therefore, $e^{i\pi}$ means starting at 1 and rotating $\pi$ (halfway around a circle) to get to $-1$.

For more details explaining each step, read this article.

score 0 · Answer 2 · answered Jun 15 '14 at 20:07

0

I believe Sal from Khanacademy gives an intuitive and simple explanation here

answered Jun 15 '14 at 20:07

Gene Yllanes

51

score 0 · Answer 3 · answered Jun 15 '14 at 20:24

If we define

$$\exp(x+iy)\equiv \exp(x)\left[\cos(y)+i\sin(y)\right]$$

then this satisfies the Cauchy-Riemann equations and is therefore a complex differentiable function. It can be shown that any complex differentiable function is analytic (i.e. it's Taylor expansion will converge to the function). So, $\exp(z)$ defined as above is an analytic function. Now, it can also be shown that if you have two analytic functions that are equal to each other on a set of points that has a limit point, then these two functions must be equal to each other everywhere. This means that any definition of the complex exponential function that reduces to exp(x) on the real line that is also analytic must be given by the above definition.

score -2 · Answer 4 · answered Jun 08 '14 at 13:52

-2

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

the $\pi$ is to some extent an arbitrary definition of the angle.

answered Jun 08 '14 at 13:52

John Fernley

1,218

1

I strongly disagree that it is at all arbitrary. If one picked a different "angle" definition, one would no longer obtain $\sin$ and $\cos$ as solutions to $y''+y=0$. – Jun 15 '14 at 20:15
1

You'd just get sin and cos with a scaling which is not far different – John Fernley Jun 17 '14 at 14:53

Why does $e^{i\pi}=-1$?

4 Answers4