Is it always the case that ${(z^a)}^b\ne z^{ab}$? I've been lead to believe elsewhere that the following is true for $z\in \mathbb{C}$ 
$${(z^a)}^b\ne z^{ab}$$  
This, of course, is not true for $z\in \mathbb{R}$  
Is this true for ${(a,b)}\in \mathbb R $ as well as $\in \mathbb C$?
Could someone please explain this in elementary terms? Why is this so?  
 A: It is a bit unclear what you are asking. The formula
$$
(z^{a})^b = z^{ab}
$$
is not true for all $x\in \mathbb{R}$ (and $a,b\in \mathbb{Q})$. See for example
$$
((-1)^2)^{1/2} = 1^{1/2} = 1 \\
(-1)^{2\cdot 1/2} = (-1)^1 = -1.
$$
Because of this, the same formula don't hold for $x\in \mathbb{C}$ since $-1\in\mathbb{C}$.
If you restrict $a$ and $b$ to the integers, then the formula holds for all $z\neq 0$ in the complex numbers.
If you restrict $a$ and $b$ to the natural numbers, then the formula holds for all complex numbers $z$. 
A: Both expressions have a countably infinite set of values when $a,b$ are real but, say, irrational. Define $L$ by
$$ - \pi < \operatorname{Im} L \leq \pi  $$ and
$$ \exp(L) = z. $$ Let's see, if $z = x + yi,$ and $L = M + N i,$ real $x,y,M,N,$ then $N y \geq 0$ and
$$ e^M = \sqrt{x^2 + y^2} \; \; \;  \mbox{AND} \; \; \; \tan N = \frac{y}{x}  $$ when $x,y \neq 0.$
The logs of $z$ are
$$ L + 2 k \pi i $$ so the values of $z^a$ are
$$ \exp (aL + 2ka \pi i). $$ The logarithms of these are, in a rather asymmetric manner,
$$ aL + 2ka \pi i + 2 n \pi i. $$
This says that all values of 
$$ (z^a)^b  $$ are
$$ \exp \left( abL + 2kab \pi i + 2n b \pi i \right)  $$ 
where the values of $z^{ab}$ are
$$ \exp \left( abL + 2kab \pi i  \right).  $$
One collection of values is a superset of the other. Further, the collection for $(z^a)^b$ and $(z^b)^a$ have considerable overlap but are different.
