Are there any proofs of Euler's Formula that do not rely on calculus?

Question

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.

Answer 1

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 $\cos(\theta)+i\sin(\theta)$.

Answer 2

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 $\sin \theta$.

Answer 3

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.

Answer 4

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.)

Answer 5

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."

Answer 6

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.

Answer 7

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

Answer 8

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 $\mathbb{Z}$ and $\mathbb{Q}$ in the usual way, e.g. $a^{-n}=\frac{1}{a^n}$ because $(a^n)(a^{-n})=a^{n-n}=a^0=1.$

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

In this paper I show that if a continuous function $\exp: \mathbb{C} \to \mathbb{C}$ satisfies the three algebraic properties I listed above, and also satisfies $\exp(\bar{z}) = \overline{\exp(z)}$ then it must be of the form $\exp(x+iy) = e^x (\cos(cy) + i\sin(cy))$ for some real constant $c$. However, these properties alone do not allow us to determine that $c=1$.

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.