We know from Euler's $e^{ix}$ formula that $e^{i\pi }=-1$. But, we also know that $e^{i2\pi }=1$, so why can't we take the two away using roots, and get $e^{i\pi }=1$ or $-1$? Similarly, why can't we say $e^{i4\pi }=1$, and use the fourth root to get 4 answers? Why can't we do this an infinite amount of times and get an infinite number of answers? I know we could say we can't do this because of contradiction, but what is the reason for the contradiction?
Brad

This is like saying, suppose x = -1, then x² = 1. If we now take roots we find x = +-1, so -1 = 1 or -1. Then we can consider xⁿ = 1 and there are n roots for x and similarly we could do this an infinite number of times and get an infinite number of answers for -1.

You just have to remember that some operations aren't reversible (ok the wording isn't right there, but I can't think of the proper term).

Regards,
Olof.

When we solve xⁿ = -1 we look for complex numbers x which satisfy the equation.

$e^{i\pi }$ is a complex number $x$ which satisfies $x^2=-1$. This does not mean all other complex numbers such numbers $x$ are equal to $e^{i\pi }$.
A similar argument might go...

Let S be the set of all people. My mum is a person therefore she is in S, therefore every person is my mum.

I see. But, then, what significance does $e^{i\pi }$ have if there are an infinite other number of ways to evaluate $e^{i\pi }$ (all of which are different)) Is it just because that solution is a more "natural" solution that we usually use it with it? Is it just easier to work with?
Brad

There is not an infinite number of ways of evaluating $e^{i\pi }$. $e^{i\pi }=-1$ is well defined (start talking about log on the complex numbers and you have a point).
There are n unique solutions to the equation xⁿ = -1. Do not confuse the two statements.

I'm not sure that's what I'm saying, though. If
$e^{i2n\pi }=1$, Then $e^{i2n\pi /2n}=e^{i\pi }=1^{1/2n}$. Of course, $-1$ is one solution to this, but it is not the only. It is the only that will appear no matter what $n$ is (unless it's 0). Perhaps I'm not understanding though (and that very could be the case)...

Sorry Brad, but this is *exactly* analogous to the example given by the first Anonymous post, i.e.

1ⁿ = 1, so 1^n/n = 1 = 1^1/n , so there are n different values for 1!

The problem is that you assume (aⁿ )^1/n = a, when it doesn't necessarily.

James.

I see now. Thanks to Olof, James, and Anon. Sorry if I sounded rude earlier-if I did it was solely because of a lack of comprehension on my part. Thanks once again,

Brad