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

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.

• It is a fact. The easiest way to prove it is via Euler's identity Commented Mar 8, 2021 at 18:36
• @DavidG.Stork No? If "an imaginary power" means a pure imaginary, rather than any complex number? I think it's just saying if $a$ is a positive real and $b$ is any real number, then $|a^{bi}|=1$. Commented Mar 8, 2021 at 19:13
• OK... if "imaginary" means "pure imaginary." Commented Mar 8, 2021 at 19:35

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

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

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

• \infty creates $\infty$ Commented Mar 8, 2021 at 19:30