The intuition of the round value of $e^{i \pi}$

Question

I can see the value of $e^{i\pi}$ is $-1$, this value is round without decimals. The $e$ value can be defined as the value (v) that the derivative function of $v^{x}$ is still $v^{x}$. And the $\pi$ value is defined as the value of circumference over diameter of whichever circle. And $i$ is $\sqrt{-1}$ everybody knows.

These 3 values $e$ and $i$ and $\pi$ have no direct relations, but $e^{i\pi}$ is a round value, what is the intuition (the simple explanation) behind this?

Answer 1

You see, there are other seemingly unlikely identities... $${\left(\sqrt{2}^{\sqrt2}\right)}^{\sqrt{2}} = 2,$$ $$\log 2 + \log 5 = 1,$$ etc.

The fact is, $\pi$ is defined such that the identity holds. Usually $\pi$ is defined as the smallest positive number such that $2\pi$ is a period of $f(x)=e^{ix}$. So $e^{i\pi}$ must be a square root of $1$, but since $2\pi$ is the smallest positive period, you can't take to positive square root. Therefore $e^{i\pi}=-1$.

To see why this is a good definition, consider doing this the other way: defining $\pi$ with circumference requires a definition of curve length, which requires the definition of improper integrals, the smallest positive period approach only requires the definition of $e^x$ as a power series sum, which only requires the definition of limits. You can also define $e^x$ as a unique solution to a differential equation, which is also straightforward. This is the definition used in Rudin's real analysis book.

Answer 2

This is just geometry. $\sin(\pi)=0$ and $\cos(\pi)=-1$ which are trivial statements if you draw the picture. But by your reasoning $\sin$ is a complicated function and $\pi$ is a complicated number while $\sin(\pi)$ is the round number $0$.

Answer 3

Really what's fundamental is not the number $e$ itself but the exponential function $x \mapsto \exp(x)$. The question is what happens when we feed this function purely imaginary numbers.

Whatever it is, it had better be on the unit circle, assuming the additive rule for $\exp$ extends to complex numbers, since $\exp(it)*\overline{\exp(it)} = \exp(0) = 1.$

So $t \mapsto \exp(it)$ is a map from a line to a circle.

If we believe that complex differentiation is to work like real differentiation, then the speed is $|d/dt \exp(it)| = |i \exp(it)| = 1$.

So the exponential function is wrapping the purely imaginary line around the unit circle at a constant speed of $1$. Since we start at $\exp(0) = 1$, by the time we're halfway around the circle we get to $\exp(\pi i) = -1$.