I actually first saw it in a cartoon. e^(pi i) = -1.

I still find it hard to believe for a simple reason. ‘i’ is an imaginary number. It has no real value.

So how could the answer be negative one?

It should be an irrational answer, shouldn’t it?

I actually first saw it in a cartoon. e^(pi i) = -1.

I still find it hard to believe for a simple reason. ‘i’ is an imaginary number. It has no real value.

So how could the answer be negative one?

It should be an irrational answer, shouldn’t it?

We have Dopers who are equipped to explain this; I am not one of them. But while we wait for someone to stop by, I can at least give you a name and an article to peruse:

BTW, before anyone else points it out, I meant *imaginary* answer. Irrational numbers are real. They just never end and never repeat (which is something else I could never understand–but that’s for a different thread ).

There is no reason that an expression containing an imaginary number cannot equal a real number. An even simpler example is i^2 = -1.

You are thinking of Gelfond’s Theorem:

If

aandbare algebraic numbers witha≠ 0, 1, andbirrational, then any value ofais a transcendental number.^{b}

This applies to e^{π} but not to e^{πi}

?

Any rational number is real, but an irrational complex number may be real or imaginary like *i*

Now that we’ve had legit answers to this question, let’s examine SMBC’s take on the issue of number types:

Me, I like the *inscrutable* ones like ‘twarch’ and ‘hun bunbred’

I often think it was a mistake for Descartes to call them ‘imaginary’ and treat them as a hack or as fictional numbers. There are a non-negligible number of people who can’t get past the ‘imaginary’ label alone. They are as real (as in ‘actual’ number, not “real numbers” in math jargon) as any number.

It’s not even an original thought on my part. Gauss believed this and said that naming them something different would have avoided a lot of confusion over the years, and he (as usual) was probably right about that.

Or i - i = 0.

Basically, a comlex number has a real part and an imaginary part. If you write an expression that causes the imaginary part to be zero, you only have the real part.

In this case, we have Euler’s formula which states that e ^ (i * theta) = cos (theta) + i sin (theta). Where theta is an angle expressed in radians. It so happens that sin(pi) is 0 and cos(pi) is -1, so you get the famous identity from the OP.

An even simpler one is the somewhat obvious e ^ (i + 0) = 1.

No, *e ^{i}* is approximately 0.54 + 0.84i

There are also good (and not-so-good, I imagine) explanations online. Here’s one of the good ones:

https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

Ack, of course that should have been e ^ (i * 0) = 1.

Agreed that “imaginary” is just a terrible word. They are simple numbers in another dimension. Real numbers are on the number line and the *other* number line that i is on runs perpendicular to it.

In fact, if you think of the “imaginary” numbers as “perpendicular” numbers, then a basic folk understanding of two-dimensional math/physics gets you to understanding the basic concept of e^(i*pi). Exponentiation is growth. If you grow in a straight line, you accelerate in that direction. What happens when you grow at a right angle to your heading? You spin in a circle.

And when you spin in a circle such that the path along that circle is pi times the radius, you end up opposite of where you were.

And my favorite SMBC comic concerning this identity (turns out there are a lot of them) is Talk mathy to me.

Gelfand’s theorem does not apply since neither e nor pi is algebraic. And if pi were algebraic so would pi*i be.

To give a quick idea of a proof, take the power series for e^x

1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + …

and replace x by ix. Collecting the real and imaginary terms and recalling the power series for sin x and cos x (written for x in radians) gives you

e^{ix} = cos x + i*sin x. Now let x = pi, whence cos x = -1 and sin x = 0.

These power series are all derived in calc 2, based on the facts that the derivative of e^x is e^x, the derivative of sin x is cos x and the derivative of cos x is -sin x (when x is in radians).

Right; the trick to apply it is that e^{π} = (–1)^{–i}

In any case, the conclusion should be that e^{π} is transcendental, while the vaguely similar-looking e^{πi} happens to be rational

My understanding (which is admittedly laughably small) is that imaginary numbers are legitimate numbers because you can apply mathematical operations to them and produce consistent answers. Which you can’t do with a zillion or a bunch or hun bundred.

Question about the comic:

If you plug -pi/2 into Eulers formula you get Cos(-pi/2) + iSin(-pi/2) which I get as -i. How is that a real number?

Little_Nemo:

iamthewalrus_3: Great_Antibob:I often think it was a mistake for Descartes to call them ‘imaginary’ and treat them as a hack or as fictional numbers.

Agreed that “imaginary” is just a terrible word. They are simple numbers in another dimension. Real numbers are on the number line and the

othernumber line that i is on runs perpendicular to it.In fact, if you think of the “imaginary” numbers as “perpendicular” numbers, then a basic folk understanding of two-dimensional math/physics gets you to understanding the basic concept of e^(i*pi). Exponentiation is growth. If you grow in a straight line, you accelerate in that direction. What happens when you grow at a right angle to your heading? You spin in a circle.

Not only is it legitimate, you can hardly avoid considering complex numbers in number theory once you consider something like the fundamental theorem of algebra, or more simply any equation like x² + 1 = 0