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

math.toronto.edu

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

#21

Post by johnaspden »

roundsquare wrote: post #18
johnaspden wrote: post #13

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…

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."

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.

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

#23

Post by tomsaffell »

johnaspden wrote: post #13

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…

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?

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

#25

Post by amalcon »

GavinB wrote: post #12

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.

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.

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

#26

Post by [deleted] »

tomsaffell wrote: post #23
johnaspden wrote: post #13

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…

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?

[deleted]

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

#27

Post by sofal »

amalcon wrote: post #25
GavinB wrote: post #12

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.

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.

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.

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

#28

Post by johnaspden »

tomsaffell wrote: post #23
johnaspden wrote: post #13

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…

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?

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.

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

#29

Post by wjy »

johnaspden wrote: post #21
roundsquare wrote: post #18

Earlier quoted context omitted.

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."

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 o…

Bravo. You have a gift for explanation. This is exactly the sort of explanation helpful for someone (like myself) who has been through all the foundations at one point or another, but sometimes forgets the big picture.

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

#30

Post by Xichekolas »

johnaspden wrote: post #21
roundsquare wrote: post #18

Earlier quoted context omitted.

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."

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 o…

For anyone that was momentarily confused by johnaspden's explanation of how (0,1) squared became (-1,0), there is excellent explanation of it here:

http://www.math.toronto.edu/mathnet/answers/imagexist.html

I'll gratuitously paste the relevant part here, but the summary is that to multiply two tuples, you simply take the cross product, as you would a binomial.

---

Remember that any collection of objects for which there is a definition of what the objects are and when two objects are equal, there is a rule for how to add two objects, there is a rule for how to multiply two objects, and these rules obey familiar arithmetic laws like commutativity, associativity, and distributivity, is, by definition, a number system.

These properties are all satisfied by complex numbers.

We have a definition of when two complex numbers are to be considered equal: they are equal if and only if they are the same pair of real numbers.

We have a rule for adding two complex numbers (which, remember, are nothing more than pairs of real numbers):

  (a,b) + (c,d) = (a+c, b+d)
and a rule for multiplying two complex numbers:

  (a,b)(c,d) = (ac-bd, ad+bc)
The rule for multiplication may look very strange, but there's nothing wrong with that; one can still verify that these rules do indeed satisfy the familiar properties of arithmetic.

Therefore, complex numbers form a number system.

Within this number system, is there an object which, when squared, gives -1? Yes. It is the pair (0,1). When you square it using the above rule of multiplication, you get

  (0,1)(0,1) = ( (0)(0) - (1)(1), (0)(1)+(1)(0) ) = (-1,0).
Strictly speaking, the complex number (-1,0) is something different from the real number -1. After all, it's a pair of real numbers, -1 and 0, not a single real number.

However, complex numbers of the form (a,0) behave identically to the way ordinary real numbers a behave. They add and multiply in exactly the same way that ordinary real numbers do:

  (a,0) + (b,0) = (a+b,0)
  (a,0)(b,0) = (ab,0)
Since numbers are just abstract concepts anyway, and since real numbers a and complex numbers of the form (a,0) are completely identical as far as their arithmetic behaviour is concerned, it is perfectly legitimate to view them as just two different representations of the same underlying concept.
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