Euler’s Identity, Part Two

Time to show you the dramatic conclusion to the story I began yesterday.

Our problem was to define the complex exponential function in a way that was consistent with everything we knew about real exponential functions. We noticed that one of the standard rules for exponents implies

$e^{x+iy}=e^xe^{iy}.$

Since we already know how to deal with the first term in that product on the right-hand side, our problem has been reduced to deciding what to do with “pure imaginary” exponents.

The key to doing that is to remember that we have a Taylor series expansion for the exponential function:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\dots,$

which is a pretty result all by itself. We have previously discussed Taylor series, so if you need a refresher course just follow the link.

Now, the cool thing about that series is that it is just as meaningful if we use a complex number in place of x. (That’s not so obvious, actually, and a rigorous course in complex variables would spend some time establishing that complex power series really are acceptable, but we will obscure those details.) We can write:

$e^{iy}=1+iy+\frac{(iy)^2}{2!}+\frac{(iy)^3}{3!}+\frac{(iy)^4}{4!}+\frac{(iy)^5}{5!}+\dots$

Let us now pause to recall some of the basic algebra of complex numbers:

$i^1=i, \phantom{xx} i^2=-1, \phantom{xx} i^3=i^2(i)=-i, \phantom{xx} i^4=i^2(i^2)=1.$

That’s very interesting, since it implies that raising i to any even power produces an integer. We will put that to good use by separating out the terms with even exponents from the series above:

$1+\frac{(iy)^2}{2!}+\frac{(iy)^4}{4!}+\frac{(iy)^6}{6!}+\frac{(iy)^8}{8!}+\dots$

Substituting in the even powers of i we saw above gives us:

$1-\frac{y^2}{2!}+\frac{y^4}{4!}-\frac{y^6}{6!}+\frac{y^8}{8!}+\dots$

If you remember your second semester calculus, that probably looks familiar. It is precisely the Taylor series for cos y.

Now let’s try the same trick for the terms with odd number exponents. We have:

$iy+\frac{(iy)^3}{3!}+\frac{(iy)^5}{5!}+\frac{(iy)^7}{7!}+\frac{(iy)^9}{9!}+\dots$

Applying our algebraic rules and then factoring out the i from every term leads to

$i \left(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\frac{y^7}{7!}+\frac{y^9}{9!}-\dots \right)$

And if that looks familiar, it is because the expression in parentheses is precisely the Taylor series for sin y.

That’s how we arrive at the definition

$e^{i \theta}=\cos \theta+ i \sin \theta,$

which leads immediately to Euler’s identity

$e^{i \pi}+1=0.$

Very nice, I’d say, and very unexpected. Of course, it gets even better when you realize that Euler’s identity is really just a special case of a more general result, that the sum of all the nth roots of unity is zero for any n. But we’ll save that for a later post.

1. #1 joemac53
December 28, 2011

Thanks for part 2. I used this relationship to tease my high school calculus students about what they could expect if they stuck with math a little longer. Of course I warned them that they may not see a recognizable number for a year or so afterwards. The kids who went the EE route all came back to say “OOOOOHHHHHH, I love that i guy!”

2. #2 NickMatzke
December 28, 2011

Thanks Jason, this is great stuff.

3. #3 Jason Rosenhouse
December 29, 2011

Nick —

joemac53 —

Your story about teasing your high school class reminds me of something that happened in graduate school. I was participating in a seminar in an area of number theory known as automorphic forms. The seminar consisted of myself, one other graduate student, and the two number theorists in the department. The other student, whose name was Karen, and I were just starting our thesis work, and we were considering going into automorphic forms. Early in the seminar Karen and I presented the proof of a difficult theorem. The proof was one of those lengthy and technical calculations that doesn’t seem to be going anywhere, until you get to the last step, when almost everything cancels and you’re left with something very elegant. I presented the proof’s finale, showed that all the bad stuff cancelled, and then said something like, “That’s really cool!” To which one of my professors replied, roughly, “That’s just the propaganda we show you when we’re trying to get students to go into this area. If you actually write a thesis you won’t be working on anything this pretty!”

As it happens Karen stuck with automorphic forms, but I went into a different area of math entirely.

4. #4 joemac53
December 29, 2011

Great story! As an undergrad I got roughed up a few times by math, and considered other majors, but I kept going back because of the challenges and beauty. One time I filled a blue book with tremendous mathematical insight, only to get it back with the note “not applicable” starting on page 2. The prof assured me that these things would happen and not to get too discouraged.
The very first time I dreamed about a math problem and solved it in my dream I did not get right up and write it down, and so lost it. After that I always got up and wrote everything down. Sometimes right, sometimes wrong. My students just shake their heads over that.
We regularly sent kids to Dartmouth, which we described as a “very mathy place”.

5. #5 Pseudonym
January 2, 2012

You can also work this out using calculus. I’m going to sacrifice some rigour for brevity, but I’m sure that any calculus student can fill in the details.

Define the function:

exp : R -> R

such that:

exp(0) = 1
D[exp(x)] = exp(x)

where D is the differential operator. (Showing that this function exists and is unique for real elements is beyond the scope of this blog comment.)

We’d like to form an algebraic extension to the complex numbers, retaining the properties:

exp : C -> C

such that:

exp(0) = 1
D[exp(x)] = exp(x)

and we would like to know what exp(ix) is for real x. We start by expanding exp(ix) two functions s and c such that:

c : R -> R
s : R -> R

exp(ix) = c(x) + i s(x)

We can expand D[exp(ix)] in two ways. One using the definition of exp and the chain rule:

D[exp(ix)] = i exp(ix) = i (c(x) + i s(x)) = – s(x) + i c(x)

The other is to take the derivative of the expansion of exp(ix):

exp(ix) = D[c(x) + i s(x)] = D[c(x)] + i D[s(x)]

Matching up real and imaginary parts gives:

D[c(x)] = -s(x)
D[s(x)] = c(x)

And since exp(0) = exp(0i) = 1, we have:

c(0) = 1
s(0) = 0

So:

exp(ix) = c(x) + i s(x)

where c(x) and s(x) are solutions to the system of coupled differential equations:

c(0) = 1
s(0) = 0
D[c(x)] = -s(x)
D[s(x)] = c(x)

Finally, show that this system of DEs has a solution which is unique.

One final thing. The complex exponential formula really shines when you go beyond complex numbers.

Let J be any mathematical object with the properties:

θJ = Jθ (i.e. J commutes with θ under multiplication)
J² = -1

Then:

e^(Jθ) = cos θ + J sin θ

So J could, be a complex unit quaternion, a matrix (e.g. a 2×2 real matrix, or i times a Pauli matrix), a Clifford algebra vector, a metric tensor or any number of other interesting things. This formula is very important in the theory of spinors, and underlies a lot of relations known to people who model Euclidean space in computers.

If J is a quantity such that:

θJ = Jθ
J² = 1

Then:

e^(Jθ) = cosh θ + J sinh θ

Finally, if J is a quantity such that:

θJ = Jθ (i.e. J commutes with θ under multiplication)
J² = 0

Then:

e^(Jθ) = 1 + Jθ

6. #6 Neil Schipper
January 3, 2012

Thanks, I enjoyed these.

Would “everyone” (authors) use definition (for e to the i*theta) as you’ve done here (as opposed to alternatives that I can’t think of)? It seems like more than a definition because it’s so motivated, so not arbitrary.

7. #7 heddle
January 5, 2012

Jason,

For completeness, explain how to handle something like [e^(2iPi)]^Pi.

Since e^(2iPi) = 1, is it 1^Pi, and does that = 1?

or, is it, using the normal rules of algebra, e^(2iPi^2) which is cos(2Pi^2) + isin(2Pi^2), which is certainly not 1.

I actually don’t know the answer to this (good thing I am only in the math department on a temporary basis), though I assume it has something to do with being on a different branch, and that the pi-th root of 1 is multivalued.

8. #8 Peter
January 6, 2012

@heddle:

Looks like you broke math! But you weren’t the first, see wikipedia, the last part about “Failure of power and logarithm identities.”
(I didn’t know that. What did I do, sleep through complex analysis?)

9. #9 heddle
January 7, 2012

Peter,

10. #10 Owlmirror
January 8, 2012

[trying to see if I can re-write comment 7 with embedded LaTeX, just because]

======

For completeness, explain how to handle something like $$[e^{2i \pi }]^ \pi$$.

Since $$e^{2i \pi } = 1$$, is it $$1^\pi$$, and does that = 1?

or, is it, using the normal rules of algebra, $$e^{2i \pi ^2}$$ which is $$\cos (2 \pi ^2) + i \sin (2 \pi ^2)$$, which is certainly not 1.

I actually don’t know the answer to this (good thing I am only in the math department on a temporary basis), though I assume it has something to do with being on a different branch, and that the π-th root of 1 is multivalued.