How to define $(-1)^{\frac24}$?

Question

Possible Duplicate:
Which step in this process allows me to erroneously conclude that $i = 1$

According to the definition of exponentials, $\displaystyle(-1)^{\frac{2}{4}}$ is equivalent to $\sqrt{1}$. However, if we take the $4$ first and the square it. The answer will be $1$. What's the problem?

Answer 1

I admit, there is a bit of apparent ambiguity here. One might say $(-1)^{(2/4)} = (-1) ^ {(1/2)} = \sqrt{-1} = i$.

But then, according to what we are accustomed to in the real numbers, we might expect to say $(-1) ^ {(2/4)} = (-1^2)^{(1/4)} = 1^{(1/4)} = 1$, in the sense that we are used to saying that the fourth root of $1$ is $1$. Or one might expect to say $(-1)^{(2/4}) = ((-1)^{(1/4)})^2 = (-i)^2 = -1$, since $i$ is a primitive 4th root of unity.

But what we're really doing here is playing with principal values. For example, $\sqrt{4} = 2$. Why isn't it $-2$? So we must not be imprecise. The idea that $b^{p/q} = (b^p )^{1/q}$ is not true in general. And that's our problem. At the end of the day, one should interpret this to mean $\sqrt{-1} = i$, without changing any sorts of order on the exponents.

Answer 2

Notice that $\sqrt{x^2} \ne x$, but instead $|x|$.