Why is e^(pi i) = -1?

http://www.math.toronto.edu/mathnet/questionCorner/epii.html

by ulvund • 17 years ago

114 47 17 years ago

math.toronto.edu

sophacles• 17 years ago

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• 17 years ago

We verify in a quick Google search: http://www.google.com/search?hl=en&safe=off&q=sin(pi...

that sin(pi) = 0.

amalcon4• 17 years ago

sin(pi)=0

I wouldn't feel ignorant or anything. The article left out that information seemingly on purpose, presumably so that people would try to figure out where the i went.

travisjeffery2• 17 years ago

The formula is:

e^(ix) = cos(x) - i sin(x)

Since sin(π) == 0, i * 0 == 0 so you have cos(π) left which is -1.

So in sum:

e^(i π) = cos(π) - i sin(π) == -1 - 0 == -1

catzaa• 17 years ago

If you draw a Sine graph, you will see that sin(x) is zero at 0, pi, 2*pi, ...

If you draw a cosine graph you will see that it is -1 at pi. Therefore in the above sum the sine part just falls away.

It is easy to see that the above is true when you recall that cosine(theta) = adjacent side/(hypotenuse). Since pi=180 degrees, adjacent will be -1 hypotenuse 1 and opposite side 0.

travisjeffery2• 17 years ago

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

tom_b• 17 years ago

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.

daredevildave• 17 years ago

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• 17 years ago

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

GavinB• 17 years ago

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.

amalcon4• 17 years ago

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.

sofal• 17 years ago

I'm not convinced that redefining Pi as 2*Pi would cause an equal amount of complications (besides confusing everyone with the change). The radius is more fundamental to a circle. The diameter is just derived from the radius. The idea of a circle with a unit radius is so widely used that it is simply called a "unit circle" with the obvious implication that the radius is what the "unit" refers to. Every time I've done math with radians, the fact that Pi is only half way around has added a subtle but noticeable disconnect. I have to force my intuition to fit the definition.

abecedarius2• 17 years ago

Yes, the alternative would cause different complications, but fewer of them. The pi-is-wrong pdf goes into it.

It was programming a lot that brought me to realize this. I got in the habit of always questioning the primitives I'm given, in the hope of simplifying the definitions built on them. It was a little boggling to notice that pi, something I'd grown up with, deserved the same kind of questioning.

mhartl5• 17 years ago

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.

johnaspden5• 17 years ago

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.

ilitirit• 17 years ago

That's a great explanation. Is this how they explain it in math classes? Because if not, then it should be. At least at a superficial level.

johnaspden5• 17 years ago

Thanks! No, I worked it out for myself while I was thinking about what the exponential of a linear operator is. The explanation at my school was the one with Taylor Series.

By the time I got to university it was just taken as a given, so nobody explained it at all.

My explanation seems to have attracted a lot of positive votes in a few minutes. I've written a more detailed version here if anyone doesn't get it:

http://johnlawrenceaspden.blogspot.com/2009/10/ei-pi-1.html

If it gets much attention I might draw the diagrams that go with it.

I'm always bewildered how many scientific types (and even some mathematicians) find complex numbers mystical. They were originally discovered by mystic methods, but Argand showed us what was really going on, and they're no more weird than 2-dimensional vectors.

Actually a lot less weird than the Reals, which really are magical and mysterious, but which everyone seems to be quite happy with!

roundsquare2• 17 years ago

Thats neat. Only thing is that (to me anyway) this takes the idea of the complex plane as being very fundamental as opposed to just something convenient. I'm not sure how to convince someone that 1 + i is the same as the coordinate (1, 1) without saying "thats just how we define it because things work out."

thwarted• 17 years ago

http://betterexplained.com/articles/a-visual-intuitive-guide...

http://betterexplained.com/articles/intuitive-arithmetic-wit...

The first link was posted here a while ago, I think.

johnaspden5• 17 years ago

The complex plane is the fundamental thing.

It's the only way to get to the complexes without doing anything spooky! Defining i to be "the square root of minus one" is about as sane as defining it to be the square root of the colour blue.

The first people who thought about it followed that approach, and were rightly scared stiff and confused by it. Even Euler made trivial mistakes.

Argand came up with the right way of thinking about it:

Take all the tuples (x,y). (Where x and y are just integers). Define on them addition and multiplication rules.

Oh look! There's a big system of these things, and embedded within it is a sub-system which works exactly the same as the integers and their rules.

Since they're exactly the same for all practical purposes, we may as well forget about the difference and say that we'll write (a,0) as a and (0,b) as ib, and (a,b) as a+ib

And although non of the pairs in that subsystem square to be (-1,0), also known as -1, there are two things that do! So now we know that both (0,1) and (0, -1), otherwise known as 1i and -1i, or i and -i, are square roots of -1.

No slight of hand or magical thinking necessary.

As it happens, we can do the same thing with the reals, to form something that embeds the reals. But the reals really are dark and mysterious and need to be brought about by a kind of magic.

tomsaffell• 17 years ago

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?

johnaspden5• 17 years ago

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.

abecedarius2• 17 years ago

Feynman gives the same explanation in his awesome Lectures On Physics, volume 1, the chapter titled "Algebra". One of the great works of 20th-century literature.

shachaf• 17 years ago

Where does e come into this specifically? That is, why does 5^(i*pi) have an imaginary part?

roundsquare2• 17 years ago

I'll give it a shot. For me, I think about e in terms of compound interest.

I.e. you have a bank account and put in a dollar. Its a great bank, they give you 100% interest. So, after a year, you have (1 + 1) = 2 dollars.

But wait, instead, they compound twice. That is, every 6 months, you get 50% interest. So now you have (1 + 1/2)^2 = 2.25 dollars.

Instead of compounding twice, they compound 4 times. So now you have (1 + 1/4)^4 = 2.44 dollars. Each time you get interest, its given in proportion to what you have in the account at the time. So, if they are compounding 4 times:

1 --> (1.25 * 1) = 1.25 --> (1.25 * 1.25) = 1.56 --> (1.25 * 1.56) = 1.95 --> (1.25 * 1.95) = 2.44.

(Please note: Rounding error. I just typed in to two places after the decimal point).

Why 2 or 4? Lets just call it compounding n times. As n gets larger, you get more money, but it doesn't go on forever...

For n times, the formula for how much money you have is (1 + 1/n)^n. What if you want to continuously compound? I.e. at every instant, growth is proportional to the current value? Well, then you would want something like (1 + 1/infinity)^infinity. But, we can't really do that, so instead we say:

lim (n --> inf) of (1 + 1/n)^n

And thats what we call e. Start putting in larger and larger values of n and you'll see that it converges to 2.718... The point is that e is pretty much defined as the number you get when you take a number starting at 1 and have it grow in proportion to itself continuously. Instead, if you use 5 instead of e, your growing too fast. In terms of johnaspden's explanation, you'd move too far around the circle, past (-1, 0).

Also, johnaspden, your explanations are truly great, really helped me understand Euler's formula.

Edit: Small change.

jrp• 17 years ago

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.

timwiseman• 17 years ago

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.

tomerico• 17 years ago

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.

tfincannon• 17 years ago

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• 17 years ago

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.

bbg• 17 years ago

http://xkcd.com/179/

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

bayareaguy• 17 years ago

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&#...

mhartl5• 17 years ago

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.

alan-crowe• 17 years ago

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)

fridgeposts• 17 years ago

A better question is how can you:

(pi/2)^2e

rameezk• 17 years ago

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

Why is e^(pi i) = -1?

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