$e^{\pi i} = -1$ is this a fact or an assumption?

Question

I think it is a fact but can someone explain why is it true intuitively?

I heard a lot of videos on youtube assuming it is the "natural" way of revolving around 0, many other explanations that does not make sense to me.

I only need a direct way of explaining why the imaginary power causes the number to become 1 in its radius (magically).

e.g. 1^i = 1

e.g. 9999^i = 1 (with some rotation)

the general intuition for exponentials is that it gives different numbers, but using an imaginary power always gives a 1 (in the radius)

this can also be proven using the formula: z = r(cos(θ) + i sin(θ)) where θ contains all the information about the number then it gets converted to an angle only without any radius other than 1.

Answer 1

If you understand Taylor expansions you can prove this property of the exponential quite easily. Assuming it is complex differentiable (see https://mathworld.wolfram.com/ComplexDifferentiable.html) you can express it as a Taylor Series (https://en.wikipedia.org/wiki/Taylor_series)

You can see that

$e^{i \theta} = \sum_{n=0}^{\infty} (i \theta)^n / n! $

Using that $i^2 = -1$ we can separate the real even terms and imaginary odd terms

$\sum_{n=0}^{\infty} (i \theta)^n / n! = \sum_{n, even}^{\infty} (-1)^{n/2} / n! + i\sum_{n, odd}^{\infty} (-1)^{(n-1)/2} / n! $

The former can be identified as the series expansion for $\cos(\theta)$ and the latter as $\sin(\theta)$. Therefore $e^{i \theta} = \cos(\theta) + i\times \sin(\theta)$ is proven.

If you were to define exponentiation to complex powers such that $9999^i = 1$, the operation $a^z$ would no longer obey the complex differentiability condition. You can define a function that behaves like normal exponentiation for real(z) but also satisfies $9999^z = 1$ if $z= i$, but then have just defined a new function that is not Holomorphic and not generally relevant to most mathematics.

Edit: \inf -> \infty

Answer 2

There is a bit of an underlying assumption here, in terms of what $e^z$ even means when $z \in \mathbb{C}$. Everything else, we can prove from that.

In one common systematic development of complex analysis, because we know from real calculus that every real number $x$ satisfies

$$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!} $$

we then define the function $\exp(z)$ similarly for $z \in \mathbb{C}$.

$$ \exp(z) = \sum_{k=0}^\infty \frac{z^k}{k!} $$

Then for a positive real number $b$, we define (assume?)

$$ b^z = \exp(z \ln b) $$

and show that it equals the ordinary meaning of $b^z$ in real numbers when $z = x + 0 i$ is a real number. This allows us to write $\exp(z) = e^z$ and use the $e^z$ notation in place of the $\exp$ function.

So determine the complex value $e^{yi}$, we plug in $yi$ as the argument to the definition of $\exp$:

$$ e^{yi} = \exp(yi) = \sum_{k=0} i^k \frac{y^k}{k!} $$

In this sum, the terms alternate between pure real and pure imaginary. Separating them out,

$$ e^{yi} = \left[\sum_{m=0}^\infty (-1)^m \frac{y^{2m}}{(2m)!} \right] + i \left[\sum_{n=0}^\infty (-1)^n \frac{y^{2n+1}}{(2n+1)!} \right] $$

Then the Taylor expansions of $\cos y$ and $\sin y$ are plainly seen, leaving Euler's formula

$$ e^{yi} = \cos y + i \sin y $$

In particular, if $y=\pi$, this is Euler's famous identity

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