Question

One of my favorite questions ever. Prove the Euler identity e^(2*pi*i) = 1

Answer 1

Different approaches encouraged =D

Answer 2

\[e^{i\pi}=-1\]

\[e^{2i\pi}=(-1)^2=1\]

Answer 3

first prove e^(i*x)=cosx+isinx by using series expansions of sin and cos
then plug in e^(pi*i) into that formula

Answer 4

That's a nice approach, Turing. Now prove it to a high schooler. LOL.

Answer 5

i^2 = -1 ... expand e using tylor's seris
use ^^ this property to prove cosx+isinx <-- also in tylor's form

and put the value of x, you will have proof.

Answer 6

I learned that in high school! (granted it was towards the end)

Answer 7

Haha really? Series expansions?

Answer 8

Taylor expansions are in AP calc part II, so yeah...

Answer 9

Well cool. =)

Answer 10

you are not in the US I presume, where they have the "AP" program

Answer 11

Haha no I am.

Answer 12

yeah .. that will work. I am sure ...
cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... + xn.cos(n.pi/2)/n! <--- these minses comes form i
sin(x) = x - x3/3! + x5/5! - x7/7! + ... + xn.sin(n.pi/2)/n! <-- also here but still there is i

Answer 13

and we know the expansion of e^x
add the above and...

Answer 14

boom we see that e^(ix)=cosx+isinx

Answer 15

I like this approach. It relies on more intuitive math.
Begin with the function f(x) = e^(ix)
f'(x) = ie^(ix)
So that means that f'(x) = i*f(x)
In the complex plane, a multiplication by i is the same as a 90 degree rotation.

Answer 16

there is a really fast proof of e^(ix)=cosx+isinx in the wiki aritcle based on differential equations, but that is clearly above high-school level

Answer 17

very nice :)
true, it's more intuitive, but treating i as a constant is only defined for real numbers in differentiation isn't it?

Answer 18

Um... no. I is a constant.

Answer 19

I've never taken complex analysis so I just wasn't sure you could use the same rules
yes, i is a constant, I guess I wanted a more rigorous proof that it works the same though it's imaginary since that since rarely found in high school calc problems

Answer 20

*since we rarely see i in high school calc

Answer 21

So what that tells me is that, in the complex plane, f'(x) is always a 90 degree rotation of f(x), so the function's derivative is always orthogonal to the function, creating a circle! =D