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There are a lot of "proofs" of the identity $e^{ix}=\cos x+i\sin x$ in textbooks, using either differential equations or power series. However, I find those proofs often misleading, because it appears to me that Euler's identity is just a beautiful definition rather than a theorem. We have good reasons to define $e^{ix}$ in this way, and those "proofs" tells us why we should define $e^{ix}$ this way, but at the end of the day, this is just a definition.

To be more precise: there are various ways to define $e^{ix}$. One very common definition is $$ e^z=1+z+\frac{z^2}{2!}+\ldots $$ If we use this as a definition, then certainly we can prove Euler's identity as a theorem. Alternatively, we can prove Euler's identity by assuming $$ \frac{de^z}{dz}=e^z, $$ but this approach requires formal treatment of complex derivatives, holomorphic functions, etc.

In the end, what we are doing is just to prove one property of $e$ from another. So, why don't we accept $e^{ix}=\cos x+i\sin x$ as a definition in schools?

Another reason why it might be seen a definition is that we are extending the definition of $e^x$ in $\mathbb R$.

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    $\begingroup$ Not an answer, and not a "school reference", but on p. 25 of Basic Complex Analysis by Jerrold E. Marsden (and on slightly later pages in later editions) one finds: But we recognize this as being simply $\cos y + i\sin y.$ So we define $e^{iy} = \cos y + \sin y.$ $\endgroup$ Commented Nov 5, 2019 at 7:04
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    $\begingroup$ I don't quite see how you would motivate that definition for students, except by showing that with this definition the other properties (expansion of $e^z$ in powers of $z$ and the differential equation) follow. But at that point you would have done the same amount of work, no? $\endgroup$ Commented Nov 5, 2019 at 8:45
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    $\begingroup$ Also, I don't think that explaining the meaning of $de^z/dz$ for complex $z$ requires anything more than explaining the meaning of it for real $z$. You don't have to mention holomorphic etc... $\endgroup$ Commented Nov 5, 2019 at 8:50
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    $\begingroup$ Explaining the meaning of $\frac{\textrm{d}f}{\textrm{d}z}$ does require really understanding what complex multiplication means, and that is actually kind of hard. Understanding the distinction between $\frac{\textrm{d}f}{\textrm{d}z}$ and $\frac{\partial f}{\partial z}$ is even harder. $\endgroup$ Commented Nov 5, 2019 at 14:55
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    $\begingroup$ @StevenGubkin it depends on what you understand by "meaning". I assume you're thinking of some geometric meaning of $df(z)/dz$? Then I agree. But the basic intuition "change in $f(z)$ per change in $z$" carries over 1-1 from real calculus, and you don't need to have a geometric understanding of complex multiplication for that. $\endgroup$ Commented Nov 5, 2019 at 20:49

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I think it is worth acknowledging that there is a huge list of equivalent definitions of the complex exponential. Any presentation should be explicit about the fact that choosing one definition over others is a choice of convenience, or taste, rather than being canonical. Showing that all of these definitions are equivalent is the real work. Here are some definitions:

  1. The unique entire function which agrees with the real exponential on the real axis (best to cover this after the identity theorem).
  2. The unique solution to the differential equation $f'(z) = f(z)$ with $f(0) = 1$ .
  3. $e^z = \displaystyle \lim_{n \to \infty} \left(1+\frac{z}{n}\right)^n$.
  4. $e^z = 1+z+z^2/2!+z^3/3! + \dots$
  5. $e^{x+iy} = e^x(\cos(y) + i\sin(y))$

There are some really beautiful interconnections between these definitions such as 3 just being Euler's Method applied to 2, 5 arising from 4, 4 arising from 1 and 2, etc. I think this exploration is worthwhile, and shows the "inevitability" of the complex exponential.

I view definition 5. as the least motivated (if it is pulled out of a hat), but the easiest to "work with". So I might use several of the other definitions as "motivation", then "derive" 5, and end up adopting 5 as the "official definition".

I also view the interplay between the complex exponential and the complex logarithm to be at the heart of an introduction to complex analysis. Every monomial other than $z^{-1}$ has an entire primitive, and so integration theory (in some sense) reduces to the study of moving around on the Riemann surface of the logarithm, which is really beautiful.

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  • $\begingroup$ You may also enjoy this answer about the real exponential (and several other answers in the same thread): matheducators.stackexchange.com/a/5689/117 $\endgroup$ Commented Nov 5, 2019 at 13:01
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    $\begingroup$ I suppose for completeness you should put a boundary condition for #2? $\endgroup$
    – kcrisman
    Commented Nov 7, 2019 at 13:20
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In my view Euler's Formula $e^{ix} = \cos(x)+ i \sin(x)$ and the resulting extension ( this is (5.) in Steven Gubkin's excellent complex answer ) of $e^{x+iy} = e^x( \cos y + i \sin y)$ is not a good definition for the exponential because it is special to the context of $\mathbb{C}$. In other algebras the interplay between the exponential function and ordinary exponential on $\mathbb{R}$ is much removed from Euler's Formula.

In contrast, if we use the series definition of $e^z = 1+z+\frac{1}{2}z^2+ \cdots $ then we can pretty easily prove all the usual features of the exponential function hold as a consequence of power series calculus (which makes sense for commutative real unital associative algebras of finite dimension). In particular,

$$ \begin{align*} e^0 &= 1\\ e^z e^w &= e^{z+w}\\ (e^z)^{-1} &= e^{-z}\\ \frac{d}{dz}e^z &= e^z \end{align*}$$

I'm pretty sure these can also be extended to the noncommutative case modulo the law of exponents (see BCH relation).

Just to make the point more explicitly, let me show what the analog of Euler's Formula is for several of my favorite low dimensional novel algebras:

  1. Hyperbolic Numbers or $\mathcal{H} = \mathbb{R} \oplus j \mathbb{R}$ where $j^2=1$ has $j$-maginary exponential (sorry, you find a better term) given by $$ e^{jx} = \cosh(x)+j \sinh(x) $$ For fun, notice $$e^{-jx}e^{jx} = 1 = ( \cosh(x)-j \sinh(x) )( \cosh(x)+j \sinh(x) ) = \cosh^2(x)-\sinh^2(x)$$
  2. Three Hyperbolic Numbers or $\mathcal{H}_3 = \mathbb{R} \oplus k \mathbb{R} \oplus k^2 \mathbb{R}$ where $k^3=1$ and $\zeta = x+ky+k^2z$ is a typical three hyperbolic variable. Here: $$ e^{kx} = l(x)+km(x)+k^2c(x) $$ where by my personal preference I call $l$ the Larry function, $m$ the Moe function and $c$ the Curly function. These can be defined by series: $$ l(x) = 1+x^3/3!+x^6/6!+\cdots $$ $$ m(x) = x+x^4/4!+x^7/7!+\cdots, $$ $$ c(x) = x^2/2+x^5/5!+x^8/8!+\cdots. $$ One of my students (Daniel Freese) noticed that there is an analog of the Pythagorean Identity for sine and cosine which constrains l,m,c as follows (my students refuse to use my three stooges terminology for some reason) $$ l^3+m^3+c^3-3lmc = 1 $$ In fact, there is such an identity for each real algebra generated by some unit $K$ such that $K^n = \pm 1$. This general result reproduces the usual result for sine, cosine and cosh and sinh as well as the 3-stooges identity above and other much uglier relations between the component functions of $e^{Kx}$.
  3. An algebraist would complain that I already gave this answer, but, they're wrong. The direct product algebra $\mathcal{A} = \mathbb{R} \times \mathbb{R}$ is funny in that $$ (a,b)(c,d) = (ac,bd) $$ makes $(1,1)$ the unity of the algebra. So, this looks a bit different, here: $$ e^{(1,0)x} = (e^x, 0) \qquad \& \qquad e^{(0,1)y} = (0,e^y) $$ So, the analog of the complex exponential is $e^{(x,y)} = (e^x,e^y)$. There is an isomorphism between $\mathcal{H}$ and $\mathcal{A}$ which can be used to connect this exponential to the one I wrote with cosh and sinh. In fact, this is another really cool aspect of the exponential; it plays nice with algebra isomorphisms.
  4. Dual Numbers here $\Gamma = \mathbb{R}\oplus \varepsilon \mathbb{R}$ where $\varepsilon^2=0$. In this case the exponential is suprisingly simple: $$ e^{\varepsilon x} = 1+ \varepsilon x $$

I've written a few papers about this sort of math and they're on the ArXiV. Perhaps some day I will find a journal where they may also find a home. My student (Nathan BeDell) also wrote a couple papers exploring what you might call precalculus for algebras.

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  • $\begingroup$ For what it's worth, this is totally a Monthly, or possibly Math. Mag., and certainly Involve, article. If these are novel (heavens knows people have written enough articles about the hyperbolic functions) you should definitely publish. $\endgroup$
    – kcrisman
    Commented Nov 7, 2019 at 13:16
  • $\begingroup$ I note no one has mentioned the exponential map ... and neither am I! $\endgroup$
    – kcrisman
    Commented Nov 7, 2019 at 13:17
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    $\begingroup$ @JamesS.Cook: The other point is that most of these algebras are special cases of Clifford algebras (for possibly degenerate quadratic forms), and this gives a uniform language for discussing and proving things about them that moreover can be found developed in standard sources. Again, the point is simply that it can be helpful to refer to this literature and to standardize the language used to refer to these objects. $\endgroup$
    – Dan Fox
    Commented Nov 13, 2019 at 10:48
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    $\begingroup$ @JamesS.Cook: My points are the following: the terminology "hyperbolic numbers" is not standard (although it does occur), which makes complicated finding references about them; basic facts about the particular algebras mentioned can all be understood as special cases, equally or even easier to prove, of general facts about some well studied classes of algebras; when dealing with exactly this sort of algebra there are far, far too many different terminologies in use; one should not ignore the huge literature about quaternion algebras (not the same as quaternions), clifford algebras, etc. $\endgroup$
    – Dan Fox
    Commented Nov 14, 2019 at 8:08
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    $\begingroup$ @JamesS.Cook: The terminological dichotomies complex/split complex, quaternions/split quaternions, octonions/split octonions are well established and correspond to the compact/split dichotomy for real forms of simple complex Lie algebras (and also the use in the context of quaternion algebras of "split" to refer to the algebra of $2 \times 2$ matrices over the base field). Although this is parallel to the elliptic/hyperbolic dichotomy, those words are used in the algebraic context with somewhat different meanings, and their use for these examples seems to me unnecessary. $\endgroup$
    – Dan Fox
    Commented Nov 14, 2019 at 8:12
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I'll put out there that it might not be helpful to have this as a definition because it wasn't discovered in that way. That is to say, Euler discovered this formula by clever manipulation of $\sin$ and $\cos$ in ways that might not quite pass muster today in terms of rigor, but were still pretty cool. According to the linked article, it used 3. in Steven's answer; apparently Newton was aware of the Taylor series for $e^x$ as early as the 1660's. To call this definition unmotivated (assuming you aren't using the other ones yet in a given course) is an understatement.

So while you could do something like you propose, and while it probably is pedagogically useful to use it as a provisional definition the first time you introduce it, I would beware saying, "And look, all the usual properties can be proved from this" rather than saying later in a course, "Now let's recall a more annoying, but good, definition of $e^x$ from a year or two ago, and show that our intuitive formula for $e^{ix}$ actually does hold using that definition."

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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. For example, 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.

You may say that you prove a#(-n) = 1/a#n by using the principle of permanence or you may say that you define a#(-n) = 1/a#n in accordance with the principle of permanence. I prefer the second formulation, but it really does not matter.

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#it = cos(ct) + i sin(ct) in which c can be any real constant. I proved that in this paper.

If you take d(e#z)/e#z = e#z or d(e#it)/dt = ie#it or e#z = 1 + z + z#2/2 + … as the property to be preserved you are compelled to c = 1 (i.e. to Euler’s formula) and all these moves are common, but I proved in the same paper that it is enough to take the differentiability of exponentiation as the property to be preserved.

We may 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 differentiability.

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  • $\begingroup$ By extending exponentiation to $\mathbb{C}$ do you mean that you want to make sense of $a^b$ for arbitrary complex numbers $a,b$? $\endgroup$ Commented Nov 6, 2019 at 9:53
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    $\begingroup$ There are plenty of real differentiable functions from $\mathbb{C} \to \mathbb{C}$ which extend the natural exponential off the real line, but do not coincide with the complex exponential. There is only one which is complex differentiable though (this follows immediately from the identity theorem, and any demonstration of the existence of such a function). $\endgroup$ Commented Nov 6, 2019 at 13:54
  • $\begingroup$ Michael Bächtold; yes $\endgroup$ Commented Nov 7, 2019 at 10:06
  • $\begingroup$ Steven Gubkin; yes, but you may prove that just by using cauchy riemann conditions applied to the general form of "exponentiation" that you are forced to by the permanence of the algebraic properties of exponentiation, cf my paper $\endgroup$ Commented Nov 7, 2019 at 10:11
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The statement $e^{ix}=\cos x+i\sin x$ can only be proven if the operation $e^z,z\in\mathbb C$ is already defined, otherwise $e^{ix}$ makes no sense. Students are likely to see $e^{ix}=\cos x+i\sin x$ before they see any other definition of exponentiating complex numbers, so the short answer to your question is yes, this identity should be treated as a definition.

What you're really showing with any "proof" of this identity is that this is the only reasonable extension of the exponential function to the complex numbers. More precisely, it is the only extension of the exponential function that preserves certain properties that we expect the exponential function should have.

This is analogous to claiming that $x^{1/2}=\sqrt{x}$. If we have only defined exponentiation $x^n$ for $n\in\mathbb N$, then $x^{1/2}$ makes no sense, and we can't "prove" it is equal to $\sqrt{x}$. What we can prove is that if we would like to extend exponentiation to rational numbers and also preserve the property $x^a\cdot x^b=x^{a+b}$, then we must have $x^{1/2}\cdot x^{1/2}=x^{1/2+1/2}=x$, and therefore we must have $x^{1/2}=\sqrt{x}$.

Similarly, if we want to extend the function $f(x)=e^x$ to complex numbers, and we want to preserve the Taylor expansion property $e^x=1+x+\frac{1}{2}x^2+\dots$, then the only possible definition turns out to be $e^{ix}=\cos x+i\sin x.$

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First, we start with those equations for sine and cosine with an infinite number of terms:

\begin{align} \sin x &= \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots \right) \\[8pt] \cos x &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \right) \\ \end{align}

We also take a formula for the exponential function also with an infinite number of terms:

$$ \exp x = e^{x} = \sum_{k = 0}^{\infty} \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

That $\exp$ is the exponential function. Both its domain and co-domain are real so far. Thus, if you feed in a real number to that function, it will spit out another real number.

Those series can be proved independently of the Euler's theorem, without relying on arbitrary-looking definitions.

But, what happens if you feed the function a multiple of $i$? Let's make a function $f$ that does that job:

$$ f(x) = \exp(ix) = e^{ix} $$

You feed in a real number to the function $f$, it gives a non-real number to the $\exp$ function and spits out whatever the $\exp$ gives as an answer. So, this proceeds to:

$$ f(x) = \exp(ix) = e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots $$

Now, considering that $i^2=-1$, that $i^3=-i$ and that $i^4=1$, this can be rewritten as:

$$ f(x) = \exp(ix) = e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots$$

Then, we split it into two sums, one without the $i$s and one with them:

$$ f(x) = \exp(ix) = e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) $$

Replacing those power series with the sine and cosine power series:

$$ f(x) = \exp(ix) = e^{ix} = \cos x + i\sin x$$

Or, simply:

$$e^{ix} = \cos x + i\sin x$$

Finally, what is $f(\pi)$?

$$\begin{align} f(\pi) &= \exp(i\pi) = e^{i\pi} = \cos \pi + i\sin \pi\\ f(\pi) &= \exp(i\pi) = e^{i\pi} = -1 + 0i\\ f(\pi) &= \exp(i\pi) = e^{i\pi} = -1\\ \end{align}$$

Or simply:

$$e^{i\pi} = -1$$

BTW, this is exactly the power series formula.

About the definition, you could argue that the formula $i = \sqrt{-1}$ is an arbitrary definition. However, the thing that actually is $\sqrt{-1}$ exists independently of what we call it, how we describe it or how we define it. So, someone just gave the name $i$ for it.

Defining something is roughly saying that you give a name to something and a description of this thing. You could define $i$ in many ways, and all of them would result in the same thing. You could define it as "let's call $i$ as the thing that multiplied by itself results in $-1$, even if it does not actually exist". Or, much more simply, you write the definition with $i = \sqrt{-1}$, where the variable to the left of the equals sign is the name that you are giving to something and the expression in the right side is the description of that thing.

However, the case of $e^{ix} = \cos x + i\sin x$ or $e^{i\pi} = -1$ is very different. They are not simple arbitrary definitions and were obtained mathematically from calculations. Defining something is very different from obtaining or inferring something.

You could argue that the equals sign is ambiguous in this case since it doesn't show clearly if the equality is the result of an inference process or simply because someone used it to give a name to something. This is why a few people use $a := b$ or $a \stackrel{\text{def}}{=} b$ for the defining equals, but most people simply don't do that or rely on the textual context to tell the difference or simply don't care at all. Thus, the case of $i = \sqrt{-1}$ is a definition, while $e^{ix} = \cos x + i\sin x$ and $e^{i\pi} = -1$ are the results of inference processes.

Finally, this all shows that those formula can be obtained without needing to resort to integral calculus nor to derivatives nor to using some previously made-up definition of how to extend the exponential function into the $\mathbb{C}$ realm. All you need to have is the power-series formula for sine, cosine and exponential, which can be proved independently beforehand, all in the $\mathbb{R}$ domain. In fact, this formula tells us exactly how to extend the exponential function into the $\mathbb{C}$ realm without having to resort on something made-up. We just fed in a non-real number instead of a real one to the exponential formula and saw if it works and what the heck was the result.

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    $\begingroup$ Hi! Can you just explain a bit more how this is answering the original question of "why not use the formula to define e^{ix}"? Probably most contributors on this thread are aware of this derivation; the question is why to choose this one as opposed to another one. $\endgroup$
    – kcrisman
    Commented Nov 6, 2019 at 21:18
  • $\begingroup$ @kcrisman Is it better now? $\endgroup$ Commented Nov 8, 2019 at 2:09
  • $\begingroup$ I suppose one could quibble about whether calculus (in Euler's sense, since those formulas are probably in *Introductio in Analysin ...") is needed to prove these things converge to what they say they converge to. But yes, this is now answering the question as written. $\endgroup$
    – kcrisman
    Commented Nov 8, 2019 at 2:17
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    $\begingroup$ How can you possibly plug in $ix$ to the real exponential? $\endgroup$
    – BCLC
    Commented Aug 11, 2022 at 9:15
  • $\begingroup$ @BCLC I think it is much more explanatory now. Too bad that my writing skills three years ago were insufficient. $\endgroup$ Commented Aug 12, 2022 at 5:16

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