The most common way I have seen Euler's formula $$ re^{i\theta} = r(\cos\theta+i\sin\theta) $$

introduced in a classroom environment is to substitute $i\theta$ into the series expansion of the exponential function, and then notice that this can be rearranged into the sum of the series expansions for $\cos\theta$ and $i\sin\theta$.

However, this requires that the students are familiar with the series expansion of these three functions, which is often taught when discussing Taylor series. This requires an understanding of derivatives.

I have seen other ways to introduce Euler's formula that rely on differential equations, however this also requires an understanding of derivatives.

I'm searching for a way to introduce Euler's formula, that does not require any calculus. The students are on an engineering course, and will have only seen algebraic manipulation, functions (including trigonometric and exponential functions), linear algebra/matrices and have just been introduced to complex numbers. They will see a fair amount of calculus (including Taylor's series) later in the course, but I would like to avoid saying "Please just accept this for now, and we will revisit it later" if possible.

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    $\begingroup$ First; is their any defintion of $e^{i\theta}$ that does not depend on calculus? YES: the defintion $e^{i\theta} = \cos\theta + i \sin\theta$. Is there any other definition? I guess "infinite series" and "limit of a sequence" and "differential equations" and "integrals" are ruled out when the OP rules out "calculus". $\endgroup$ Jan 15 '20 at 13:13
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    $\begingroup$ Have a look at the brilliant video at youtube.com/watch?v=v0YEaeIClKY $\endgroup$
    – Sasha K
    Jan 15 '20 at 22:05
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    $\begingroup$ How do we even define $\sin, \cos$ without a notion of a limit? $\endgroup$ Jan 16 '20 at 0:06
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    $\begingroup$ @MatthewNajmon That definition works intuitively, because we have an idea of what it means to move along a circle. But a rigorous definition along those lines requires a notion of arclength, which requires a notion of limit. $\endgroup$ Jan 16 '20 at 0:59
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    $\begingroup$ @OrangeDog Any rigorous definition of $\exp$, $\sin$, $\cos$, etc. requires something beyond first-order geometry. This is because the real numbers and the algebraic numbers have the same first-order properties (they are both real closed fields). Therefore, there is no way to define transcendental functions such as $\cos$ using only first-order statements about finite numbers of real variables. $\endgroup$ Jan 16 '20 at 16:49

Can you define $e^x$ for real $x$ without calculus? You need to take a limit, I think.

If you allow "informal limit reasoning", then there is such an argument.

One way to define $e^x$ for real $x$ is by the limit $(1+\frac{x}{n})^n$ as $n$ grows without bound. This is often motivated with a compound interest story.

Substituting $i\theta$ in for $x$ into this definition we have the limit of $(1+\frac{i\theta}{n})^n$ as n grows without bound. $1+\frac{i\theta}{n}$ is geometrically very close to $\cos(\frac{\theta}{n})+i\sin(\frac{\theta}{n})$ (wishy washy, but can be made precise. This precision is "differential calculus"). By De Moivre's formula, $e^{i\theta}$ should be very close to $\cos(\theta)+i\sin(\theta)$.

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    $\begingroup$ You can define exponentiation without calculus (not sure though if you can for a complex exponent). With this, the OP might be reformulated as "prove that there exists such a constant $e$ that Euler's formula holds true". This might not require calculus. $\endgroup$
    – Ruslan
    Jan 15 '20 at 22:05
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    $\begingroup$ I think I can build something around this. The students have been introduced to $e^x$ via the classic "compound interest story" so that can be used again to get a basic demonstration of Euler's formula. A more rigorous derivation can be shown later in the course when they get to calculus. $\endgroup$ Jan 16 '20 at 9:11
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    $\begingroup$ @Ruslan You can define exponentiation of integer and rational exponents without calculus. But real exponents? I think inf/sup or limit is necessary, even if reals are introduced by axioms (instead of constructing them, which requires inf/sup/lim). $\endgroup$
    – Pablo H
    Jan 16 '20 at 14:33
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    $\begingroup$ "Compound rotation" would be the intuition needed here. $\endgroup$
    – J.G.
    Jul 16 '20 at 14:24

There is a proof of de Moivre's formula $$ (\cos\theta + i \sin\theta) ^n = \cos(n\theta) + i \sin(n\theta), \qquad n \in \mathbb Z $$ by induction (for $n > 0$) and symmetry. Maybe that is the best we can do without calculus.

Some textbooks (not assuming calculus) use a notation $\mathrm{cis}\;\theta$ meaning $\cos\theta + i\sin\theta$ and do all the calculations with it. The addition formula for $\mathrm{cis}\; \theta$ combines the two addition formulas for $\cos\theta$ and $\sin \theta$.

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    $\begingroup$ de Moivre's formula is a relationship between one combination of trigonometric functions and a different combination of trigonometric functions. It does not link trigonometric functions to exponential functions without additional steps. If you can make those extra steps without relying on calculus, or simply asserting $re^{i\theta} = r(\cos\theta + i\sin\theta)$ as an a priori fact, then I would be interested. I think I can see the direction you are going, but you would need to put in those extra steps before I could accept this answer. $\endgroup$ Jan 15 '20 at 13:57

The title of the question asks for a proof, but the actual question seems to be more about providing some motivation. For students at this level, I would aim more for motivation that they will understand rather than a proof that they will just see as a mysterious magic trick.

Here's how I motivate it. I have the students work through $i^0$, $i^1$, $i^2$, ... with me, while I stand at the board and plot each point in the complex plane as we go along. With college students who have had a previous, brief intro to complex numbers, I find that many are surprised by the fact that it goes around in a circle, and they volunteer that it's cool.

Optionally, do a second example in which you plot $(\sqrt i)^n$.

Once they've seen this, it should be clear that you can take any number $b$ on the unit circle, and when you do $b^r$, where $r$ is real, you're going around the unit circle with an angle proportional to $r$. Clearly it would be convenient to find the base such that the constant of proportionality is 1.

For students at this level, who have not even officially learned limits, I would just jump from that to stating Euler's formula without proof. If this is a precalculus class, then as preparation for calculus I think it would be valuable to have them see an informal discussion of a limit like $\lim_{n\rightarrow\infty} (1+x/n)^n=e^x$, but I think that would be valuable in the more familiar context with real numbers. If there are one or two students in the class who could really absorb something like this as applied to complex numbers, then it might be appropriate to provide them with some kind of written treatment that they could read if they're interested, at their leisure when they have time to think it over.

  • $\begingroup$ When you say "take any number b on the unit circle," is b the angle theta? For example, could I take b to be pi/4? $\endgroup$
    – Kara
    Feb 12 '20 at 22:16
  • $\begingroup$ @Kara The complex number $b$ would be $\cos(\pi/4)+i\sin(\pi/4)$. $\endgroup$ Jul 14 '20 at 18:58

You could show that, if $A^{ix} = f(x) + i g(y)$ is defined in a way that it satisfies $A^{ix+iy} = A^{ix} A^{iy}$ and other standard properties of the exponential function, then $f$ and $g$ must satisfy the addition and subtraction formulas for cosine and sine, so they have to be $\cos(kx)$ and $\sin(kx)$ for some $k$.

To preserve properties of exponential when extending to imaginary exponents:
- $A^0i = 1$ implies $f(0) + ig(0) == 1 + 0i$ implies $f(0) = 1$, $g(0) = 0$

  • $A^{x+y} = A^x * A^y$ ==>
    -- $A^{i(x+y)} = A^{ix+iy} = A^{ix} * A^{iy}$
    -- $f(x+y) + ig(x+y) = (f(x) + ig(x))*(f(y)+ig(y))$
    -- $f(x+y) + ig(x+y) = f(x)f(y) + if(x)g(y) + ig(x)f(y) - g(x)g(y)$
    -- $f(x+y) + ig(x+y) = (f(x)f(y) - g(x)g(y)) + i(f(x)g(y) + g(x)f(y))$
    -- $f(x+y) == f(x)f(y) - g(x)g(y)$ and $g(x+y) = f(x)g(y) + g(x)f(y)$

Also, the distinction between $i$ and $-i$ is arbitrary, so when extending a real-valued function $h$ to a complex-valued one, it is natural that $h(\overline{z}) = \overline{h(z)}$

  • $\overline{A^{ix}} = A^\overline{ix}$
    -- $f(x) - ig(x) = A^{-ix}$
    -- $f(x) - ig(x) = f(-x) + ig(-x)$
    -- $f(x) == f(-x)$, $g(x) = -g(-x)$

  • $f(x + (-x)) = f(0)$
    -- $f(x)f(-x) - g(x)g(-x) = 1$
    -- $f(x)f(x) + g(x)g(x) = 1$

But showing that for $A = e$, the functions taking radian measure work probably isn't possible. (As others mentioned, defining $e$ without calculus is problematic.)


I think calculus is not technically needed if you are merely interested in algebraic features of the imaginary exponential. In particular, let us note that: $$ \cos (\theta + \beta) = \cos (\theta) \cos (\beta) - \sin( \theta) \sin(\beta) $$ and $$ \sin (\theta + \beta) = \sin(\theta) \cos (\beta)+ \sin( \beta) \cos (\theta). $$ These may both be derived without calculus (provided you use a geometric definition of sine and cosine). Then, since $i^2=-1$ and $a+ib=c+id$ iff $a=c$ and $b=d$ we can package the equations together as one wonderful complex equation: \begin{align} \cos(\theta + \beta)+i\sin(\theta + \beta) &= \cos \theta \cos \beta +i^2 \sin \theta \sin\beta +i \left( \sin \theta \cos \beta+ \sin \beta \cos \theta \right) \\ &= (\cos \theta + i \sin \theta)(\cos \beta +i \sin \beta) \end{align} Thus, if we simply introduce the notation $$ e^{i \theta} = \cos \theta + i \sin \theta $$ then the law of trigonometry above is simply written: $$ e^{ i (\theta+ \beta)} = e^{ i\theta}e^{ i \beta} $$ Moreover, we can solve $e^{i \theta} = \cos \theta + i \sin \theta$ and $e^{-i \theta} = \cos (-\theta) + i \sin (-\theta) = \cos \theta - i \sin \theta$ and obtain from adding and subtracting equations the elegant algebraic formulations of sine and cosine: $$ \cos \theta = \frac{1}{2}\left( e^{i \theta}+e^{-i\theta} \right) \qquad \& \qquad \sin \theta = \frac{1}{2i}\left( e^{i \theta}-e^{-i\theta} \right) $$ from which we may explicitly and algebraically derive nearly any trig. identity. Alternatively, introduce $e^{i\theta}$ as a notation for $(\cos \theta , \sin \theta)$ where $e^{i(\theta+\beta)} = e^{i \theta}e^{i\beta}$ can be derived by direct geometric reasoning.

Fine, but... why use the notation "e" for $e^{i \theta}$ ? I would simply say that calculus is best used to explain this because the exponential is an object which properly belongs to calculus. Furthermore, I use the notation $e^{i \theta}$ because it is standard notation which you will continue using once you know more math. Since math is a language it is good to learn the words even if we cannot stomach the etymology at a given juncture.


I believe Euler's identity can be reached via De Moivre's Formula:

$$\cos(nx)+i\sin(nx)=\left( \cos(x)+i\sin(x)\right)^n$$

(I am not finding a clear exposition of this route, as often Euler's identity is used to prove De Moivre's, whereas here we're seeking the reverse.)

Wikipedia says, "The truth of de Moivre's theorem can be established by using mathematical induction."


I think that Mathologer's explanation does not require calculus. He uses the concept of limits and relies on geometric interpretation of complex multiplication. It is the simplest explanation that I am aware of. https://www.youtube.com/watch?v=-dhHrg-KbJ0

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    $\begingroup$ If you are taking limits, then you are (in my opinion) doing calculus. In any event, you answer is pretty much just a link. Answers on SE should be self-contained---please edit your answer to explain or give an overview of what is discussed if one clicks the link. $\endgroup$ Jul 14 '20 at 13:03
  • $\begingroup$ This is a nice video! It is an explanation of the same approach I suggest in the accepted answer. However, he does even more hand-waving towards then end, because he doesn't make the observation that the angle of $1+i\frac{\pi}{n}$ is approximately $\frac{\pi}{n}$ (this is what I am doing with my sines and cosines). So his explanation of why it wraps exactly halfway around the circle is incomplete. Very nice visuals though! $\endgroup$ Jul 14 '20 at 13:37

A guiding principle for operations extension is the principle of permanence: a definition of an operation should be extended from a restricted domain to a wider one in such a way as to conserve the crucial properties of the operation.

The crucial algebraic properties of exponentiation: a#(b+c)=(a#b)(a#c), (a#b)#c=a#(bc) and (ab)#c=(a#c)(b#c) compels us to define the exponentiation in Z and Q in the usual way, e.g. a#(-n)=1/a#n because (a#n)(a#(-n))=a#(n-n)=a#0=1.

To extend the exponentiation to R you should count continuity of exponentiation among its crucial properties and then define it on R in such a way as to conserve the crucial properties: its algebraic properties and the continuity. What you get is the usual exponentiation on R.

If you want to extend it to C, these properties are not enough. They almost compel you to Euler’s formula. Namely, they compel you to the formula e#(ix) = cos(cx) + i sin(cx) in which c can be any real constant. I proved that in


The fact that c=1 iff that function is (complex) differentiable is a trivial consequence of Cauchy-Riemann conditions.

To conclude: by the principle of permanence of the algebraic properties and the continuity we are almost compelled to Euler’s formula. We are definitely compelled to it if we extend the principle to the (complex) differentiability.

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    $\begingroup$ Link-only answers are generally discouraged on stackexchange. $\endgroup$
    – user507
    Jan 24 '20 at 0:06
  • $\begingroup$ I deleted links, Is it better now? $\endgroup$ Jan 26 '20 at 12:37
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    $\begingroup$ No, I think it is worse. You should include in your post the definition of "the principle of permanence". And, if possible, a short indication of the proof. But also include the link. $\endgroup$ Jan 26 '20 at 13:21
  • $\begingroup$ @ZvonimirSikic The map $a+ix \mapsto e^a(\cos(cx) + i\sin(cx))$ is real differentiable (considered as a map from $\mathbb{R}^2 \to \mathbb{R}^2$). It is true that the only value of $c$ which makes this map complex differentiable is $c=1$, but your paper is surely not the first one to show this. In fact, by a basic theorem (the identity theorem) there is at most one entire function extending $e^x$ to the complex plane. So it was already known, for more general reasons, that there is only one way to extend real exponentiation to complex if you want the result to be holomorphic. $\endgroup$ Jan 27 '20 at 21:00
  • $\begingroup$ As far as I know my paper is the first one to derive e*a (cos cx + isin cx) from the principle of permanence (and to introduce that function). That is my contribution. The fact that c=1 iff that function is (complex) differentiable is a trivial consequence of Cauchy-Riemann conditions. $\endgroup$ Feb 4 '20 at 12:49

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