eiy = cos(y) + isin(y)

i is defined as sqrt(-1)

Euler's identity, given above, is a wonderful and mysterious result. The identity binds geometry with algebra and often simplifies the mathematics of physics and engineering (see phasor for an example). In some sense Euler's identity is more a definition than a result--one could define eiy in other ways. However, the chosen definition is clearly the most useful one.

Complex functions have very nice mathematical properties if they are analytic. In fact, complex analysis is based on analyticity over regions of the complex plane, allowing for isolated singularities. Given the solid structure inherent in analytic functions, it is desirable to define ez, where z = x+iy, in such a way that it is analytic over the entire complex plane. Taylor's Theorem tells us that if a function f(z) is analytic over the whole complex plane (entire), then it can be written as Σ1/n! f(n)(0)zn. Since Taylor's Theorem must hold for y = 0, the power series in z must be the same as the real-valued power series in x. In other words, if we want ez to be entire then it must be given by Σ1/n! zn.

Given this reasonable definition for ez we can obtain eiy by plugging iy into the power series. Miraculously, the result is the sum of the power series of cos(y) and isin(y)!

A special case of Euler's identity, when y = π, is a very famous equation since it is composed of five fundamental, seemingly unrelated mathematical constants. This equation is e + 1 = 0.

e + 1 = 0

Not a definition at all, but rather a proof of the fundamental interconnectedness of all things and the unreasonable effectiveness of mathematics, this is something that still blows my mind. After I read it in a book, I told it to my high school math teacher, but she didn't believe me. It blew my father's mind when I told it to him, but of course he demanded proof. I couldn't pull that off until college, but when I could, this is the proof that I gave him, and will now give to you. Keep in mind that this uses calculus, so if you haven't had any exposure to Taylor Series, this won't mean much to you. Alright, let's begin...

First, some notation: ! means "factorial".

        x! = "x factorial" = 1 * 2 * 3 * 4 * ... * (x-1) * x
        4! = "4 factorial"
           = 1*2*3*4 
           = 24
For the real proof, you have to remember that notation, and you have to remember Taylor Expansions. It's all about Taylor Expansions. (Actually, these are special-case Taylor series, and are called Maclaurin Series, but only hard-core math geeks care about the difference) Taylor expansions use derivatives around f(0) and polynomials to approximate a function. As they use more derivatives ("as n goes to infinity") they become exact approximations. When I put ... below, it means "this series continues on to infinity".

Here is the Taylor expansion of sin(x):

        sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ...
And here is cos(x):
        cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - ...
And here is the Taylor expansion of ex:
        ex = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! + ...
Now let's look at the taylor expansion of eix, while remembering the fact that i2 = -1:
        e^(ix) = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! + (ix)^4/4! + ...
               = 1 + ix/1! + i^2*x^2/2! + i^3*x^3/3! + i^4*x^4/4! + ...
               = 1 + ix + -1*x^2/2! + -1*i*x^3/3! + x^4/4! + ...
               = 1 + ix - x^2/2! - ix^3/3! + x^4/4! + ix^5/5! - x^6/6! + ...
Now, notice that this is just the Taylor expansion of cos(x) plus i times the Taylor expansion of sin(x) (this fact is the amazing one!). Therefore:
        e^(ix) = cos(x) + i*sin(x)
From there it is a short step to:
        e^(i*pi) = cos(pi) + i*sin(pi) 
                 = -1 + i*0 
                 = -1
Proving, for once and for all, that:
        e^(i*pi) = -1
        e^(i*pi) + 1 = 0
Whoah. This last equation is definitely the preferred form, because it includes all of the really important numbers: 1, e, π, and 0. The fact that this is even true at all has led to people referring to it as god's equation, and has led many a good mathematician astray into fields of philosophical wonder.

vruba says Re Euler's identity: he was a Swiss math guy named "Euler", you big silly.

Not too long ago, I attended a lecture given by William Dunham about Leonhard Euler. Prior to this lecture, I thought the only way to derive Euler's identity used Taylor expansions, or that there was some complex method for deriving it which I was unable to comprehend at the time. Yet there is an absolutely beautiful proof for this identity which Dunham showed us, and which I will now show my fellow E2ers.

To understand the following proof for Euler's identity, ei x = cos(x) + i sin(x), you need to know the following tidbits of The Calculus:


FIRST: We have the following integral:

∫ dz / √(1 + z2)

NEXT: Since we are mathematicians, and want to save time, we look this up in a table of integrals. (N.B. if you feel so compelled, you could evaluate the integral using trig substitution.)

∫ dz / √(1 + z2) = ln (z + √(1 + z2))

MORE NEXT: We'll perform a simple variable substitution (obviously, i = √-1 here) by letting z = i y:

z = i y
dz = i dy

∫ dz / √(1 + z2) = ln ( z + √(1 + z2) )

∫ i dy / √(1 + (i y)2) = ln ( i y + √(1 + (i y)2) )
i ∫ dy / √(1 - y2) = ln( i y + √(1 - y2) )

THENNILY: we perform trigonometric substitution by letting y = sin(x)

y = sin(x)
dy = cos(x) dx
N.B.: 1 - sin2(x) = cos2(x)

i ∫ dy / √(1 - y2) = ln( i y + √(1 - y2) )

i ∫ cos(x) dx / √(1 - sin2(x)) = ln( i sin(x) + √(1 - sin2(x)) )
i ∫ cos(x) dx / √(cos2(x)) = ln ( i sin(x) + √(cos2(x)) )
i ∫ cos(x) dx / cos(x) = ln ( i sin(x) + cos(x) )
i ∫ dx = ln ( i sin(x) + cos(x) )
i x = ln ( i sin(x) + cos(x) )

We're almost there....
FINALLY: Raise e to the power of both sides:

ei x = eln( i sin(x) + cos(x) )
ei x = i sin(x) + cos(x)

quod erat demonstrandum, beeyatch.
There you have it: a proof of Euler's identity that doesn't involve Taylor Series. It's so purty.


  • William Dunham's lecture, given on 11 May, 2005
  • Wikipedia (For the table of integrals)
  • My internalized Calculus notes

This proof has been brought to you by the letter e, the number i, and the person Leonhard Euler.

* You'll notice that I'm missing the constant of integration. That's because we don't really need it here. Besides, Leonhard Euler didn't use it! Why should we? (N.B. this excuse does not work with your Calculus teacher, and it doesn't work with College Board. Don't try this at home, kids.)

** Yes, we can treat i as a constant here. In Euler's time, i was considered a bit of a taboo number (hence the somewhat derogatory term, "imaginary number"). Euler made i hip, though. He felt that if you couldn't treat i as a constant, you'd be destroying the foundation of mathematics.

† You don't need to worry about absolute value signs when you're dealing with these numbers. When you have an integral whose answer involves an ln (usually from u substitution), putting on absolute value signs really doesn't change much, except make it "prettier" when dealing with real numbers only. Since we're not dealing with real numbers, they have no need. Besides, the only difference between ln(x) and ln(-x) is a constant factor. Especially where integrals are involved, it's just like treating it as a constant of integration, which can "drop out," anyway.

I scribbled up some kind of proof some minutes ago.. I like it because it doesnt bring up
so many unnatural subjects.

If one has a linear homogeneous ode with characteristic equation (r-i)^2, then its solutions are (C+Dx)e^ix
The equation can be written u''-2iu'+u=0, where ' denotes differentiation with respect to x.
Then we just happen to try if (i sin(x) + cos(x)) satisfies the equation.. and it does.

Furthermore: all particular solutions to the equations can be written (C+Dx)e^ix,
so isinx+cosx = (C+Dx)e^ix, letting x=0 gives C=1, and by raising everything to 2 (don't know how to say that :P)

we get (isinx+cosx)^2=(cos^2x - sin^x) + 2icosxsinx = isin2x+cos2x = (1+2Dx)e^2ix = (1+2Dx+D^2 x^2)e^2ix, so obviously
D=0, and so we have isinx+cosx=e^ix

This proof (if it is a proof :P) uses complete solution to lin. hom. odes, common trig formulas, and a bit of simple algebra. I hope its valid.

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