Confused about complex numbers I am confused about something:
\begin{eqnarray}
    (e^{2 i \pi})^{0.5} = (e^{2 i \pi \cdot 0.5})= e^{i \pi}=-1
\end{eqnarray}
but
\begin{eqnarray}
     e^{2 i \pi}=1~ and~ 1^{0.5}=1 
\end{eqnarray}
where is my mistake??
thanks
 A: When dealing with complex numbers, you no longer interchange radicals with fractional exponents, in the same way you did with positive real numbers.
If $x$ is real, $\sqrt[3]x$ denotes the principal cube root of $x$, and has only one value. In this way $8^{1/3}$ and $\sqrt[3]{8}$ mean the same thing.
But if we are doing algebra with complex numbers, and we wrote $5^{1/3}$, we would mean any solution to $z^3 - 5 = 0$.
So $1^{1/2}$ is defined as any solution to $z^2 - 1 = 0$. $z = 1$ and $z = -1$ are both solutions to this.
A: Exponentation $x^y$ can be unambiguously defined whenever $x$ is a positive real number, and $y$ is any complex number. The value is (per definition if you like) equal to $\exp(y\ln(x))$ where $\exp:\Bbb C\to\Bbb C$ is the usual exponential function, which is well defined and well behaved. [Writing it using $e^z$ would give a circular definition here.]  But the rule $x^{yz}=(x^y)^z$ is only valid when $y$ is real; this is more strict than requiring $x^y$ to be real (as the question illustrates), which is the condition necessary for $(x^y)^z$ to be defined in the first place.
The proof that $x^{yz}=(x^y)^z$ when $x>0$ and $y$ are real is simple:
$$
  x^{yz}\stackrel{\rm def}=\exp(yz\ln(x))
= \exp(z\ln(x^y))\stackrel{\rm fed}=(x^y)^z
\qquad(x,y\in\Bbb R, x>0, z\in\Bbb C).
$$
This also shows that that a situation with $x^{yz}\neq(x^y)^z$ where both sides are defined but $y\notin\Bbb R$ must boil down to a similar but more basic situation involving $\ln$. Indeed the rule $y\ln(x)=\ln(x^y)$ is not valid for $y\notin\Bbb R$ even if the arguments to $\ln$ are both real:
$$\def\ii{\mathbf i}
  2\pi\ii=2\pi\ii\ln(e)\neq\ln(e^{2\pi\ii})=\ln(1)=0.
$$

Curiously, the failure of $x^{yz}=(x^y)^z$ when $y\notin\Bbb R$ does not dissuade people to employ this very rule when trying to "define" such exressions as $\ii\,^\ii$. I find such attempts truly pathetic. There is no point in trying to define exponentiation beyond the case of positive real base, or integer exponent; one gains nothing but confusion.
A: Here, let me try and make an analogy that you might be able to understand simply. As stated in the comments, 1 has two square roots: 1 and -1. See why this is important below.
$$((-1)^{2})^{0.5} = (-1)^{(2 \times 0.5)} = (-1)^1 = -1$$ 
OR
$$((-1)^{2})^{0.5} = 1^{0.5} = \sqrt{1} = 1$$ 
Do you get it?
A: It's exactly as if you would write:

I am confused about something: 
  $$((-1)^2)^{0.5} = (-1)^{2\cdot 0.5} = (-1)^1 = -1$$
  but
  $$(-1)^2=1 \text{ and } 1^{0.5} = 1$$
  where is my mistake??

