Paradox or simplicity?

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I do not believe that there are any contradictions here.
The beauty of math is such that concepts should reinforce each other. There should be no contradictions otherwise math as we know is not true. That means we cannot even add competently.

As an addition, we can view e^(iΘ) as cisΘ=cosΘ+isinΘ.
Then a property is (cisΘ)^k = ciskΘ. this should work for any real number k.

To countercheck your statements:
(1) e^( iπ ) = -1 = cisπ
(2) w = e^( 2πi/3 ) = cis 2π/3
(3) ( e^( iπ ) )^( 2/3 ) =(cisπ)^2/3 = cis2π/3 = w. TRUE
(4) -1^( 2/3 ) =(cisπ)^2/3=cis2π/3
=cos2π/3+isin2π/3= ½ ( -1 + i√3 ). TRUE
(5) w^( 3/2 ) = e^( 2πi/3 )^( 3/2 ) =(cis 2π/3)^3/2 = cisπ = -1.
QED.
Hope it helps.

Edit: w is not the imaginary part of the eisenstein integers.
The imaginary parts are (i*n√3)/2 where n are integers.
w is the primitive complex cuberoot of unity.

Addendum: Due to the fundamental theorem of algebra. Any polynomial of degree n has exactly n complex roots.
That is why w, w², and 1 are the 3 complex cuberoots of 1.

Next EDIT:
WOW! THIS IS SO MUCH FUN. INTELLECTUAL DISCUSSION!!

ZANTI and Hipguyrockin gives sensible arguments.
But remember: The field of complex numbers is completely closed. That is, you cannot create a bigger extension field for the complex field. Whenever you have a polynomial with even complex coefficients all its answers can be located within the field of complex numbers.

Zanti, there is no such conclusion implied (x^n=y^n → x = n) and that is not an algebraic claim anyway.

Hip..., recall that for 1^¼ the number i is its primitive root.

I actually have to research to give credence to my claim:
http://en.wikipedia.org/wiki/Complex_numbers_exponential#complex_powers_of_complex_numbers
See complex powers of complex numbers

Going back to what I said before, any complex number can be written as a+bi. -1^( 2/3 ) is a complex number.
Note the general rule for complex numbers:
z^w = e^(w log z) where z,w are complex numbers and log z means the principal value of the complex logarithm. z must not be zero.

Now, -1^( 2/3 ) = e^((2/3) log(-1)) and log(-1) is the complex number πi. Thus -1^( 2/3 ) = e^(2πi/3) = (cis 2π/3).

I still hold to my original answers.
There is just a caution, there is a failure of power and logarithm identities. See the website provided.

Thank you for a stimulating intellectual discourse....Show more

w is not an Eisenstein integer. It is only part of an Eisenstein integer. w is a complex cube root of 1.

What you're getting at is the fact that:

cube root( w^3 ) = 1, apparently.

That's true. It's a cube root of 1. So:

cuberoot ( w^3) =

cuberoot( 1 ) = 1

BUT that's only the CONVENTIONAL cube root of 1. There are several. Another cube root of 1 is w. The problem is that taking square/cube/whatever roots of complex numbers does not give unique results....Show more

Well, I think some of your conclusions look reasonable, but it looks invalid to say -1^( 2/3) = ½ (-1 + i√3). The numbers 1, -1, and ½ (-1 + i√3) are all solutions to the equation x^6 = 1, but they are different numbers, of course. You can't say x^6 = y^6 implies x = y any more than you can say x² = y² implies x = y.

Also, you have to be careful when you start taking roots of complex numbers. For example, you can just as easily say that e^(3πi) = -1, therefore e^(πi) = e^(3πi), therefore πi = 3πi, therefore 1 = 3... hmm, that can't be right :)...Show more

The reason for this apparent paradox is simple and will be apparent when you consider this:

(-1)^(1/2) is meaningless in the real number system. but what if I write (-1)^(1/2)=(-1)^(2/4)=(1)^(1/4) = 1

the reason for this absurd (and in fact wrong) result is that in case of the real system, for the function a^x , we assume that a>0 because only then will raising it to a real power be meaningful in the real field.

however, by generalizing to the complex field, we obtain the facility to raise any complex number to another complex number. but we pay a price - the result is no longer "THE RESULT" - a complex number raised to a complex power has several results.

the multiplicity of results makes the identity (a^m)^n = a^(mn) invalid in many cases of complex numbers....Show more

I think you're breaking some ad hoc rules dealing with the treatment of complex numbers....Show more