When I first saw the identity $e^{i\pi} + 1 = 0$, I was awestruck at how $5$ fundamental mathematical constants could combine so elegantly, finding invariance in all their numerical chaos. It was one of the first moments that made me appreciate the inherent beauty of mathematics.

It was not long, however, before this superficial facade broke (or so it seemed) when I learned that $e^{i\pi}$ was not as closely linked to the traditional concept of exponentiation (repeated multiplication) as I had initially presumed. As I attempted to parse through the complex beauty of this deceptive creature, I settled on the idea that this was an extension of repeated multiplication into the complex plane, where one can think of the complex exponentiation as a vector rotating about a circle. This was satisfactory but not enough $-$ it felt like I had clarified something in my mind but had left behind that nudging childlike fascination that I felt when I first saw the identity. I felt like I had stumbled upon a relatively much more boring interpretation of this identity which was, unfortunately, the correct one. I asked myself this question then and I still ask myself this question today:

Is Euler's identity really that beautiful or is it notationally contrived to be beautiful?

Note that I am not questioning the validity of the identity $-$ I am well aware of the proof that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$.

I was particularly prompted to ask this question today as I came across a podcast with Grant Sanderson (Math YouTube legend aka 3Blue1Brown) and Lex Fridman (Cool MIT guy aka the host) where Grant says that $e^{x}$ is notational abuse.

The podcast on YouTube titled Grant Sanderson: 3Blue1Brown and the Beauty of Mathematics | Lex Fridman Podcast #64 is highly enlightening (at least for me) and I would recommend you watch the whole thing. But anyways, for the relevance of this post, the timestamp that is important is $3$:$48$ to $10$:$25$ and I'd love to hear your thoughts on what Grant says during this period of the video.

Now I know a bit about Math.SE and I know that I'm treading slightly towards subjectivity because of vague notions of beauty so I'll try to suggest some simple criteria (I can't find a better word sorry, I know this kinda gives off exam vibes) for answering questions:

  • Please focus your answers on the notational merits of denoting whatever the heaven the identity is doing with exponentiation.
  • Please do not start a war about whether this is the most beautiful equation in mathematics or not.

If you've made it so far into the post (firstly, thank you), do consider smashing that upvote button and leaving a comm- (just kidding, do as you please :)).

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    $\begingroup$ One trouble with your post is that there's not any mathematical question here. This is a site for answerable mathematical questions. From the tour: Not all questions work well in our format. Avoid questions that are primarily opinion-based, or that are likely to generate discussion rather than answers. $\endgroup$
    – Lee Mosher
    May 10, 2022 at 14:21
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    $\begingroup$ The point is that $e^z$ is defined in a more intrinsic way: it is equal to $\sum_{n \geq 0}\frac{z^n}{n!}$ for any complex number $z$. You may say that this "defines" $\cos z$ and $\sin z$ via the equation $e^{iz} = \cos z + i\sin z$ (and its complex conjugasion), but then the beautiful thing is that the $\cos$ and $\sin$ defined in this way coincide with those defined by ratio of sides of triangle. $\endgroup$
    – WhatsUp
    May 10, 2022 at 14:24
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    $\begingroup$ For purposes of suggesting how to focus down to a more answerable question, let me ask: Are you primarily asking how Euler's formula is related to the traditional concept of integer valued exponentiation as repeated multiplication? $\endgroup$
    – Lee Mosher
    May 10, 2022 at 14:24
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    $\begingroup$ Au contraire, I would say that it exposes what exponentiation does. $\endgroup$
    – user619894
    May 10, 2022 at 14:31
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    $\begingroup$ @Random It sounds like you don't have a problem with the definition of $e^x$ for real $x$. Thing is that the only way to extend that definition to complex $x$ is the way it is defined now. If you don't find that natural, consider that the only alternatives would be to either not define $e^x$ at all for $x \in \mathbb C \setminus \mathbb R$, or define it in a way that would be incompatible with, and break the basic properties of, the exponential $e^x$ over reals. $\endgroup$
    – dxiv
    May 10, 2022 at 16:41

1 Answer 1


There are many layers to the equation $e^{it}=\cos(t)+i\sin(t).$ We need to understand what the symbol $e^{it}$ is intended to denote, and why it is associated with exponentiation, even if it may not necessarily be exponentiation.

Exponentiation as Repeated Multiplication

At the most basic level, $x^n$ is defined as repeated multiplication, and so $n$ must be a nonnegative integer. $x^0=1$ is a consequence of the idea that the empty product is $1,$ and then $x^{n+1}=x^n\cdot{x}.$ The idea behind this definition is that it allows you to say that $x^0=1,$ $x^1=1\cdot{x}=x,$ $x^2=x\cdot{x},$ $x^3=(x\cdot{x})\cdot{x}$, etc. This is mostly a notation tool: we do this, because writing repeated products explicit as products would be extremely impractical, and repeated products show up way too often in mathematics to bother with such impracticality.

An Important Theorem About Repeated Multiplication

Using the definition above, you can use the axiom of induction to prove the statement $x^{m+n}=x^m\cdot{x^n},$ at least if you are working with real numbers, complex numbers, matrices of real numbers, or objects of that sort. This is an extremely important equation: it is the equation that makes the notion of exponentiation as repeated multiplication useful in mathematics. In some sense, this equation defines what exponentiation is as an algebraic concept.

A More Sophisticated Definition of Exponentiation

Because of this theorem, we can characterize exponential sequences differently than we typically would with the basic definition. If you let $f_x(n)=x^n,$ then $f_x$ satisfies the functional equation $f_x(m+n)=f_x(m)f_x(n),$ where $m, n$ are nonnegative integers. In fact, we can say this: an exponential sequence is any nonzero sequence $f$ satisfying the functional equation $f(m+n)=f(m)f(n).$ An exponential sequence of base $b$ is an exponential sequence satisfying the condition $f(1)=b.$ There are major advantages to taking this more technical and abstract approach: it allows for generalizations, and it allows us to talk about exponential functions and power functions without having to appeal to repeated multiplication.

Exponential Functions in Real Analysis

In the definition above, we required that the functions satisfying $f(m+n)=f(m)f(n)$ be sequences: functions whose domain is $\mathbb{Z}_{\geq0}$. But what if we insist that the domain be $\mathbb{R}$ instead? So a natural question to ask is, can we find functions such that they satisfy $f(x+y)=f(x)f(y)$ for all real $x,y$? Moreover, can we find such functions that are also continuous? The answer to this question is Yes. The nonzero continuous functions satisfying the above equation are called exponential functions, and the unique continuous function satisfying the above equation, and additionally satisfies $f(1)=b,$ is called the exponential function of base $b.$ Something that you will find is that while, for exponential sequences, any nonzero base is allowed, in the case where the domain is $\mathbb{R},$ the only bases allowed are $b\gt0.$ This is because, if, for example, $f(1)=-1,$ then there is no $f$ satisfying the condition that also satisfies the equation. But this caveat aside, this is equation is how we define exponential functions when we allowed the "exponent" to vary continuously, rather than discretely, at least in the case that the codomain is also $\mathbb{R}.$ It gets more complicated when the codomain is something else. What is nice about this definition is that, from this definition, all the important properties of the exponential function can be easily proven: you can prove monotonicity, invertibility, and you can prove that $f(x)\gt0,$ and that $f'=f.$ And once you can prove $f'=f,$ every other property I have not mentioned here becomes essentially trivial. Given this definition, it makes some sense why people use the notation $b^x,$ even though my opinion is that it is not sufficiently justified. This is a lot like conflating $n!$ with $\Gamma(x+1).$ The latter is an extension of the former, but we use different notations because they still ultimately represent different mathematical objects, and the distinction is important. With $e^n$ versus $e^x,$ it works the same way. This is why I prefer using the notation $\exp$ instead of $e^x,$ personally, but I can understand why people use the notation $e^x,$ even if it is not sufficiently justified.

Trigonometric Functions Defined Analytically

The above paragraphs explain how we analytically reason about extending the notion of exponentiation from being repeated multiplication to some idea that is more abstract. The trigonometric functions are more complicated to define this way, though. Historically, trigonometric functions were first introduced as ratios of sides of similar triangles as "measured in degrees," and each ratio corresponds uniquely to some angle. However, this has very limited applications, and when you define the trigonometric functions as functions, instead of geometric ratios, you need to forget the idea of angles, and instead focus on functional equations, or using analysis, to define these functions. There are multiple ways to do this. Many people define the trigonometric functions as functions satisfying the differential equations $x'(t)=-y(t),$ $y'(t)=x(t),$ with conditions $x(0)=1$ and $y(0)=0.$ Another way to define them is to define them is as the unique continuous functions $x$ and $y$ satisfying $x(t)^2+y(t)^2=1$ such that they have half-period $\pi$ and $x(0)=1$ and $y(0)=0$ while $x(\pi)=-1$ and $x(\pi)=0.$ This latter definition is motivated by the unit circle and geometric considerations. Yet another way to define them is to actually start with the inverse trigonometric functions instead, typically by defining $$\arcsin(t)=\int_0^t\frac1{\sqrt{1-u^2}}\,\mathrm{d}u,$$ and this can be motivated for algebraic reasons, analytic reasons, as well as geometric reasons. Once you have any of these definitions, you can compute many of the simple properties you expect $\sin$ and $\cos$ to have, and you can prove their differentiability and analyticity on $\mathbb{R}.$

Maclaurin Series Expansions and Extensions to the Complex Numbers

With all three definitions now well-defined as functions of real numbers, in a precise sense, you can do analysis with them. In particular, you can find their derivatives, and furthermore, you can prove that they have Maclaurin series expansions that will converge to those functions everywhere. These Maclaurin series expansions are what we use to extend these functions to the complex numbers: we redefine them so that the Maclaurin series becomes the primary concept, and the functions are just the specific Maclaurin series we choose them to be.

Euler's Formula

Building on everything established above, it may be worth asking "can we use the Maclaurin series expansion of $\exp$ to find a simple expression for $\exp(it)$?" It turns out you can, and when you do, you end up proving Euler's formula: $e^{it}=\cos(t)+i\sin(t),$ or as I would write it, $\exp(it)=\cos(t)+i\sin(t).$ But notice this: at this stage, having gotten to a place where the theorem becomes provable, the function $\exp$ is no longer something that a person would recognize as actually being "exponentiation," given all the layers of redefinition that we had to go through, with each layer getting further and further away from what most people recognize as exponentiation, which is just the basic notion of repeated exponentiation. Indeed, I would agree that the equation $e^{it}+1=0$ is misleading: because while the equation is technically true, if you adopt the convention of notation that $e^{it}=\exp(it),$ it leads people to make many conclusions that are not true, and it makes the connection between $e$ and $\pi$ seem much more direct than it actually is. See, at the stage at which we are working with complex numbers, we are not even working with a general notion of exponential functions with multiple bases satisfying a functional equation, since the notion is not particularly well-defined within the complex numbers, but instead, we are working with an extension of the natural exponential function to the complex number via a series expansion. This is part of what makes the notation $e^{it}$ to be misleading: it makes it seem like the notion of a "base" is still valid at this stage, when in reality, we have adopted a redefinition of the function that is no longer suited for this idea. However, there is still a real, genuine, legitimate sense in which $e$ and $i\pi$ are related here: we know that $\exp(1)=e,$ while on the other hand, $i\pi$ is the half-period of $\exp,$ and indeed, $\exp(i\pi)=-1.$ In fact, $\exp(2\pi{i})=1.$ So one could even get cheeky and say $\exp[\exp(2\pi{i})]=e.$


So, is the exponential function that appears in Euler's formula truly a form of exponentiation? Well, it is a function that is various layers removed from what most people would call exponentiation, and so the notation is indeed a bit misleading, but ultimately, this function is based on, and is related to, the notion of exponentiation as being repeated multiplication, even if the two notions are not directly identical. And despite the notation being misleading, the formula does still imply an important and seemingly fundamental relationship between $e$ and $\pi{i},$ even if less direct, and more importantly, it relates three important functions from real analysis, and it reveals them as being really all "the same function" in the context of complex analysis and beyond, in some sense.

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    $\begingroup$ +1 $\tau \lambda \alpha \eta \kappa \; \gamma \phi \mu$ $\endgroup$
    – user905694
    May 10, 2022 at 22:05

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