How do complex number exponents actually work?

Question

I know Euler's formula and how to take complex exponents, but in it it's $e$ to an imaginary angle, not a number, it seems. From my understanding pi itself is not an angle, but $\pi$ radians is. And since cosine can only take in an angle, or at least a representation of one, and the variable is the same everywhere in Euler's formula, the exponent should be an angle and not a number. The question is then raised could I then say $e^{180i}=-1$? Surely this has a single numerical value though, right?

I've been bothered by this for a long time. Am I wrong or how can I take an exponent with a complex number?

Edit for the duplicate: I understand how to take them and how they're supposed work and be justifyed/proved, but raising something to the power of an angle which is what I thought was meant to be happening made me question how they actually worked that could allow this.

Answer 1

This is more an answer to the title of your question. A possible (and perhaps the standard) way of properly defining things is as follows. See any reasonable rigorous textbook on analysis for the proofs.

First define the exponential function $\exp : \mathbb{C} \rightarrow \mathbb{C}$, by the absolutely convergent series $\exp(z) := \sum_{n=0}^{\infty}{ \frac{z^n}{n!}}$. It is easy to prove that $\exp$, restricted to the real line, takes real values. A bit more work shows that $\exp$, restricted to the real line, is strictly incresing and everywhere $>0$ and that $\exp(\mathbb{R}) = (0, \infty)$. It follows that there is a bijetive function $\log{}:(0, \infty) \rightarrow \mathbb{R}$, the inverse of $x \mapsto \exp(x)$ from $\mathbb{R} \rightarrow (0, \infty)$. The number $e$ is defined as $e: = \exp(1)$. For any positive real number $a$ and any complex number $z$, one defines $a^z$ by $a^z := \exp(\log(a) z)$. Hence, by definition, $e^z = \exp(z)$ (because $\log{(e)}=1$, by definition) and this makes sense for all complex numbers $z$.

One defines the functions $\cos, \sin : \mathbb{C} \rightarrow \mathbb{R}$ by $\cos(x) := \frac{1}{2}(e^{ix}+ e^{-ix})$ and $\sin(x) := \frac{1}{2i}(e^{ix}- e^{-ix})$. It is a (rather non-trivial) Theorem that there is a unique smallest, strictly positive zero $p$ of the function $\cos$, restricted to the real line. One defines the number $\pi$ by $ \pi := 2p$. It can be shown that the number $\pi$ and the functions $\sin$ and $\cos$ have the "familiar properties". Notice that the identity $e^{ix} = \cos(x) + i \sin(x)$ for $x \in \mathbb{R}$ follows directly from the definitions. As does $e^{i \pi} = -1$. It always strikes me that people find these two identities so amazing.

Answer 2

There can be a bunch of definition of number $e$. Let's stick with this definition: $$ \frac{d}{dt} e^t = e^t. $$

So we want to find what number can be $z(t)=e^{it} =x(t)+iy(t)$: $$ z'(t)=x'(t) + iy'(t)=ie^{it} = ix(t) - y(t),\\ x'(t) = -y(t),\qquad y'(t)=x(t),\\ x''(t) = -y'(t)=-x(t) $$

Finally, $x''(t)+x(t)=0$ and $x(0)=1$, $x'(0)=-y(0)=0$ (since $e^{0i}=e^0=1$), thus $x(t)=\cos t$ as the only function that suffice the equation. Then $y=x'(t)=\sin t$.

So, we have found what number is $e^{it}=\cos t+i\sin t$

Answer 3

Angular Units and Trigonometric Functions

When we talk about arguments to trigonometric functions, there are at least $2$ common angular units: degrees ($360$ to a full rotation) and radians ($2\pi$ to a full rotation). A number of calculators also support gradians ($400$ to a full rotation), which are only used in some countries and usually only in certain occupations (e.g. surveying, mining, geology).

In mathematics, we usually use radians because when angles are measured in radians, we have $$ \lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=1\tag1 $$ That is, for small angles, $\sin(x)\sim\tan(x)\sim x$. The actual ordering for $|x|\lt\frac\pi2$ is $$ \frac{\sin(x)}{x}\le1\le\frac{\tan(x)}{x}\tag2 $$ Furthermore, when $x$ is in radians, we have the nice series $$ \sin(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}\tag3 $$ and the value of $$ \arctan(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}\tag4 $$ is in radians.

Radians are also natural because an arc which subtends $x$ radians on a circle of radius $r$ has length $rx$.

So when we talk about angles, and don't mention the units, we assume radians.

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The Exponential of Imaginary Numbers

For $x\in\mathbb{R}$, we can write $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag5 $$ so it seems reasonable to write $$ e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n\tag6 $$ Multiplication by $1+\frac{ix}n$ increases the absolute value so minimally, that even when repeated $n$ times, it is insignificant as $n\to\infty$. However, multiplication by $1+\frac{ix}n$ rotates a number on the unit circle by a distance of $\frac xn$ counter-clockwise along the circle. When this is repeated $n$ times, it rotates a number on the unit circle by a distance of $x$.

Thus, $e^{ix}$ is a point on the unit circle whose counter-clockwise distance from $1+0i$ is $x$. This is why we use radians when saying $$ e^{ix}=\cos(x)+i\sin(x)\tag7 $$ To see a more detailed explanation of $(7)$, see this answer.

Answer 4

In calculus, the trigonometric functions are not dealing with angles but with real numbers. Implicitly, the numbers are taken to be radians, so that $\cos\pi=-1,\sin\pi=0,$ which is compatible with Eulers' famous formula

$$e^{i\pi}=\cos\pi+i\sin\pi=-1.$$

These function are "natural", in the sense that they resemble their derivatives:

$$(e^{i\pi})'=ie^{i\pi}=(\cos x+i\sin x)'=i(\cos x+i\sin x).$$

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You can very well define functions assuming arguments in degrees and write for instance $\cos_d180=-1,\sin_d180=0$. But these functions are a kind of "historical mistake" and do not enjoy the above derivative property, as an extra scaling factor appears.

Answer 5

Just a little note that I hope can be of some use. In order to avoid confusion when dealing with angles and unit of measures of them, one can directly define the trigonometric functions starting with the arc lenght measurement.

Consider the semicircle of equation $$f(x) = \sqrt{1-x^2}, \ \ x\in [-1,1].$$ It is relatively easy to show that the arc length between point $(y, f(y))$ and point $(1,0)$ can be computed with the (improper) integral $$A(y) = \int_y^1\frac{1}{\sqrt{1-x^2}}dx, \ \ y\in [-1, 1].$$ Here the unit of measure is clearly the same as the one used for determining abscissae and ordinates.

$A(y)$ is continuous in $[-1,1]$, differentiable in $(-1,1)$, and strictly decreasing. So it has a well defined inverse function $$A^{-1}(x)= \cos(x)$$ with domain $[0, \pi]$, where, by definition $A(-1) = \pi$. The rest of the cosine function can be defined using appropriate shifts and periodicity.

Answer 6

A number of people are giving fairly complex answers (no pun intended) about exponentiation, but I want to address a much simpler misunderstanding in your answer, which should hopefully clear up some of the issues.

And since cosine can only take in an angle

No, the trig functions definitely take numbers. They're related to the circle, in that they're cyclic, and the "width" of their repetition is exactly 2π, which happens to be the circumference of a circle when you measure it in radians. But the argument to trig functions aren't angles except by convention.

The question is then raised could I then say e¹⁸⁰ⁱ=−1?

180 isn't equal to π, any more than it's equal to 50 just because 180 degrees is 50% of circle. 180 degrees is equal to π radians, but since the exponentiation operator also takes numbers, not angles, that equivalence between angle units isn't relevant.