Does the Equation (e2Pi i)i = 1 Have a Geometric Meaning?

[abhi2005singh]

(e^{2Pi i})ⁱ = 1
(e^{2Pi i})ⁱ = e^{(2Pi i)*i} = e ^-2Pi2

 

[Hurkyl]

I see two problems:

(1) Your arithmetic doesn't make sense to me
(2) You are assuming exponentiation is single-valued

(compare with the argument 1 = sqrt(1) = -1)

 

[HallsofIvy]

(e^{2Pi i})ⁱ = 1
(e^{2Pi i})ⁱ = e^{(2Pi i)*i} = e ^-2Pi2

Multiplying \pi i by i does not square the \pi!

What you should have is (e^{2\pi i})^i= e^{-2\pi}. Now, what reason do you have to say that that is equal to 1? (1^x= 1 only for x real.)

 

[Max™]

e to i times pi is a half circle back to -1, e to i times two pi would circle the rest of the way around to 1.

http://www.wolframalpha.com/input/?i=e^(2pi*i)

Whereas e^((2pi*i)i) is e^(-2pi).

http://www.wolframalpha.com/input/?i=e^((2pi*i)*i)

 

[abhi2005singh]

e to i times pi is a half circle back to -1, e to i times two pi would circle the rest of the way around to 1.

Is it necessary to invoke geometrical meaning. Can't simple algebra prove the point? Anyways, I did not understand how this explains the problem.

Multiplying LaTeX Code: \\pi i by i does not square the LaTeX Code: \\pi !

What you should have is LaTeX Code: (e^{2\\pi i})^i= e^{-2\\pi} . Now, what reason do you have to say that that is equal to 1? (LaTeX Code: 1^x= 1 only for x real.)

I got my mistake. Thanks for pointing out.
1^x= 1 only for x real. This does not seem to be correct. I tried using Mathematica also. It gave result 1.

You are assuming exponentiation is single-valued
(compare with the argument 1 = sqrt(1) = -1)

I fail to understand why u didn't give one multi-valued exponentiation function to drive home ur point.

 

[abhi2005singh]

From whatever analysis I have done for this problem, I have reached to the following conclusion.
For an expression like x^a*b, u cannot write it as (x^a)^b when a and/or b are complex numbers. The order of evaluation matters. (x^a)^b can be different than (x^b)^a. The product "a*b" HAS to be evaluated so as not to arrive at contradictions and to make evaluation unique. Such order of evaluation is frequent in many branches of mathematics. I will give an example.

Fallacy: e^-2Pi = exp[2Pi*i²] = [exp(2Pi*i)]ⁱ = 1ⁱ = 1.

Consider the following which will highlight that order of evaluation matters.

(eⁱ)^2Pi*i = 0.00186744 = e^(-2Pi2)


(e^2Pi*i)ⁱ = 1.

You get two different results in the two cases. However, the result will be unique if the product in the exponent is determined FIRST. The expressions

(eⁱ)^2Pi*i = 0.00186744 = e^(-2Pi2)

and

(e^2Pi*i)ⁱ = 1

both are correct as here we have explicitly mentioned the order of evaluation. The problems starts only when we write
x^(a*b) = (x^a)^b = (x^b)^a.

Here we have CHANGED the order of evaluation and then we are expecting for getting same answers.

The concept of
x^a*b = (x^a)^b = (x^b)^a
is borrowed from the properties of real no.s which complex no.s are not bound to follow.


Above statements are based on my own analysis of the problem and hence may or may not be correct.

Any corrections/improvements/comments/suggestions please.

 

[Max™]

The geometric meaning pops to mind for me because I first learned about the identity written as e^ix = cos x + i sin x, which for pi gives -1.

Like that.