# Where is the fault here

1. Jun 28, 2010

### abhi2005singh

(e2Pi i)i = 1
(e2Pi i)i = e(2Pi i)*i = e -2Pi2

2. Jun 28, 2010

### Hurkyl

Staff Emeritus
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)

3. Jun 28, 2010

### HallsofIvy

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.)

4. Jun 28, 2010

### Max™

5. Jun 29, 2010

### abhi2005singh

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

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.

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

6. Jun 29, 2010

### abhi2005singh

From whatever analysis I have done for this problem, I have reached to the following conclusion.
For an expression like xa*b, u cannot write it as (xa)b when a and/or b are complex numbers. The order of evaluation matters. (xa)b can be different than (xb)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)

(e2Pi*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

(e2Pi*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) = (xa)b = (xb)a.

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

The concept of
xa*b = (xa)b = (xb)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.