

Why is e^(pi i) = -1? - ulvund
http://www.math.toronto.edu/mathnet/questionCorner/epii.html

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johnaspden
Imagine you're a complex number, which is just a type of 2-vector.

Exponentiation is to do with growth at a speed which is a multiple of how big
you are already.

i is the multiplication which turns you through ninety degrees.

If you grow in a direction which is at right angles to yourself, you turn
rather than increasing in magnitude.

Pi is how long it takes you to turn through a half circle.

So if you grow at right angles to yourself for time pi, you are pointing the
opposite way.

~~~
tomsaffell
This is a good explanation of why the result is negative and why it has zero
imaginary component - ie why it is _pointing the opposite way._. But it
doesn't explain why the result is _unit one_ in magnitude / length. Can you
extend it do that that?

~~~
johnaspden
Thanks! Yes, when you grow in a direction at right angles to yourself, you
don't increase in magnitude. So the turning arrow is the same length at the
end as it was at the beginning.

This is actually a slightly tricky point, because it requires the direction of
growth to change as you grow. If you grow by a finite amount all at once, then
you do get slightly longer. So you might expect an outward spiral.

But a point moving round in a circle is always moving at right angles to the
radius connecting it to the centre.

And in the same way, an arrow whose tip is moving at right angles to its shaft
isn't extending. If the length is changing, then the direction isn't 90
degrees.

To reason about this properly I think you need some sort of theory of
infinitesimals and continuous motion.

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GavinB
Pi is Wrong: <http://www.math.utah.edu/~palais/pi.pdf>

It should be e^(pi i) = 1, but pi was unfortunately defined at half the
appropriate value in the 17th century.

~~~
amalcon
Pi is defined as the circumference of a circle of unit diameter. This isn't
somehow less correct than the alternative of making that circle one of unit
radius. It causes some annoying extra factors (as the PDF points out), but the
alternative would cause different complications. The area of a circle, for
example, would be r^2*pi/2.

~~~
mhartl
_The area of a circle, for example, would be r^2 pi/2._

This is actually my favorite example of why pi is wrong—it's the "exception"
that proves the rule. To see why, set _τ_ = _C_ / _r_ = 2 pi, and then
consider the following chart of common quadratic forms:

    
    
      integral of u       1/2   u^2
      kinetic energy      1/2 m v^2
      distance fallen     1/2 g t^2
      spring energy       1/2 k x^2
      triangular area     1/2   b h
      circular area       1/2 τ r^2
    

We see that, far from causing "different complications", using the right
circle constant brings the area of a circle into a more natural form. The 1/2
in the formula for circular area is actually a _missing_ factor; using tau in
place of pi restores it.

N.B. I made a previous comment along these lines, but put it in the wrong
place. If you're still a "pi is wrong" skeptic, considering the improvement in
radian angle measure may yet convince you:

The explanations mapping complex exponentiation to rotations are basically
right. In this context, it's worth noting that the conventional choice of
circle constant is off by two. We should be using tau = _τ_ = _C_ / _r_ as the
circle constant, rather than pi = _π_ = _C_ / _D_. Read _τ_ as "turn" and all
those radian angle measures suddenly make sense. Ninety degrees? Instead of
the confusing _π_ /2 we have 90° = _τ_ /4 = one quarter turn. And so on: 60° =
_τ_ /6 = one sixth of a turn, 180° = _τ_ /2 = one half turn, etc.

In these terms, Euler's formula would be recast as

 _e_ ^( _i_ _τ_ ) = 1

That is, the exponential of the imaginary unit _i_ times the circle constant
_τ_ is unity: one full rotation.

~~~
amalcon
All of those (except the circle) have the 1/2 because they're integrals of
something linear. While it's true that area and integral are closely related
(the latter being a special case of the former), a circle is clearly not
linear.

~~~
johnaspden
As a circle expands, its area grows proportional to its circumference. The
circumference is proportional to the radius.

So you're getting the area by integrating a linear thing.

~~~
mhartl
_As a circle expands, its area grows proportional to its circumference. The
circumference is proportional to the radius._

Exactly. In symbols, this reads _dA_ / _dr_ = _C_ ∝ _r_. Setting _τ_ = _C_ /
_r_ = 6.2831853…, we have _dA_ = _C_ _dr_ = _τ_ _r_ _dr_ => _A_ = ½ _τ_ _r_ ².
Q.E.D.

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travisjeffery
The article didn't have any mention of Euler, but this is typically called
Euler's Formula and is considered one of the most beautiful formulas to
Mathematicians.

Here's some more info if you want to find out some info, applications and
such: <http://en.wikipedia.org/wiki/Eulers_formula>

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daredevildave
Write this as e^(iπ) + 1 = 0 and you have my favourite equation.

It relates all the fundamental mathematical numbers: e, i, π, 1 and 0.

~~~
JBiserkov
... using the fundamental equivalence relation =, and the operations +, * & ^

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alan-crowe
You remember those half-angle formulae

sin 2A = 2 sin A cos A

cos 2A = (cos A)^2 - (sin A)^2

If you are happy using sine of small x is x and cosine of small x is 1 as your
base cases, you can write sine and cosine as mutually recursive functions:

    
    
             (defun sin (z)
               (if (small z)
                   z
                   (* 2 
                      (sin (half z))
                      (cos (half z)))))
    
             (defun cos (z)
               (if (small z)
                   1
                   (- (square (cos (half z)))
                      (square (sin (half z))))))
    

(sin pi) => 6.167817939221069d-7 quite close to zero (cos pi) =>
-1.0012055113842453d0 you can get this much closer to -1 by using 1-x^2/2 as
your base case.

It had never occurred to me to do exponential the same way, as

    
    
             (defun exp (z)
               (if (small z)
                   (+ 1 z)
                   (square (exp (half z)))))
    

So it is a bit of a shock to try it and see it work just fine for complex
numbers

(exp (complex 0 pi)) => #C(-1.0012055113842453d0 6.167817939221069d-7)

------
mhartl
The explanations mapping complex exponentiation to rotations are basically
right. In this context, it's worth noting that the conventional choice of
circle constant is off by two. We should be using tau = _τ = C/r_ as the
circle constant, rather than pi = _π = C/D_. Read _τ_ as "turn" and all those
radian angle measures suddenly make sense. Ninety degrees? Instead of the
confusing _π_ /2 we have 90° = _τ_ /4 = one quarter turn. And so on: 60° = _τ_
/6 = one sixth of a turn, 180° = _τ_ /2 = one half turn, etc.

In these terms, Euler's formula would be recast as

 _e_ ^( _i_ _τ_ ) = 1

That is, the exponential of the imaginary unit _i_ times the circle constant
_τ_ is unity: one full rotation.

------
bayareaguy
I enjoyed Lakoff and Núñez's section on this in their book _Where Mathematics
Comes From_ [1]. The relevent pages are available at Google Books[2].

1- <http://en.wikipedia.org/wiki/Where_Mathematics_Comes_From>

2- [http://books.google.com/books?id=YXv6SEjTNKsC&pg=PA432&#...</a>

------
tom_b
For those of you interested in a longer look at this, check out "E, the Story
of a Number" by Eli Maor.

Yes, I read when I should probably be hacking.

------
jrp
I like Feynman's description, where he actually used 10^x first, just noting
that 10^x for small real x was 1 + ln(10)*x (approximating from the
derivative), assuming that this worked for small complex values too, and then
extended to larger values by squaring.

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sophacles
So to broadcast my math ignorance to all:

Why does cos(pi) + i sin(pi) == -1 ??

Why does i disappear in this?

Edit: Thanks to the replies below.

~~~
rottencupcakes
We verify in a quick Google search:
[http://www.google.com/search?hl=en&safe=off&q=sin(pi...](http://www.google.com/search?hl=en&safe=off&q=sin\(pi\)&aq=f&oq=&aqi=g10)

that sin(pi) = 0.

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bbg
<http://xkcd.com/179/>

I'm sure almost everyone has seen this anyway, but why not post.

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timwiseman
So many texts on complex analysis simply define e^{i \theta} = \cos \theta + i
\sin \theta without ever explaining how. This provides a good introduction to
the reason behind with only minimal recourse to Calculus.

Another good description is discussed in "Visual complex Analysis" by Needham.

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tomerico
I have known it as the celebrities formula, because if you write it this way:
e^(pi i) + 1 = 0

It contains all the "celebrities" of the math world.

It's quite cool actually.

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tfincannon
Pages like this will be so much better when we have MathML in the browser and
can stop using fuzzy images for mathematical formulas.

~~~
btilly
If you control the website just use <http://www.math.union.edu/~dpvc/jsMath/>
and you're fine. Any users that want can install the fonts and will get a
better experience.

Given how much harder MathML is to author, the jsMath solution is always going
to be more popular among random math folks. Particularly since they're already
used to writing TeX.

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fridgeposts
A better question is how can you:

(pi/2)^2e

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rameezk
umm i m no smart ass.. but i think e^(pi i) = -1 cuz LHS = cos PI + isin PI =
-1 .... :O

