What is the difference between $x^{1/2}$ and $\sqrt{x}$? As far as I know, $x^{1/2} = \sqrt{x}$. On the other hand, $\sqrt{i^4} = \sqrt{1} = 1$ and $(i^4)^{1/2} = i ^ {4/2} = i^2 = -1$. Does this mean that $\sqrt{x} \neq x^{1/2}$?
 A: You use the exponent law $\left( a^b \right)^c = a^{bc}$ in the following step:

$\color{red}{(i^4)^{1/2} = i ^ {4/2}} = i^2 = -1$ 

but that rule, while true for real numbers and a positive base $a$, does not hold in general. You can get in trouble with negative bases and more generally with complex numbers. 
For more details and some examples, you can check Failure of power and logarithm identities.
A: That is a question of definitions. Already in the reals, you have $\sqrt{(-1)^2}$, that could be either one or minus one. In the reals, we solve this problem by defining that the square root of $x > 0$ should always be the unique positive number $y >0$, solving $y^2 = x$. However, we don't have an ordering -- i.e. a concept of positive or negative -- in the complex field. Thus here we run into problems.
It is still true that $\sqrt{x}$ and $x^{1/2}$ mean the same, however, you have to define what a square root is.
As the complex field is algebraically closed, the equation $y^2 = x$ will always have two different solutions for $x \neq 0$ and thus you need a way to distinguish them to talk about a well-defined root. There are ways to do it, if you are interested in it you might want to study complex analysis.
Either way, you should remember that the rules for computing with these functions are not always true, i.e. the rule $(a^x)^y = a^{xy}$ only holds when both $x$ and $y$ are integers.
A: The problem is the application of the usual power laws to complex numbers.
The principal complex power is defined as
$$
z^\alpha=e^{\alpha \,\text{Log}z}
$$
with Log(z) the principal logarithm (note the capitals in Log and Arg).
The principal square root of $i^4$ is then
$$
(i^4)^{1/2}=e^{1/2 (\text{Log}\,i^4)}
=e^{1/2(\log|i^4|+i\text{Arg}(i^4))}
=e^{1/2(\log|1|+i0))}
=e^0=1
$$
Both square roots of $i^4$, $\pm1$, follow from the multi-valued nature of the non-principal log function (note the lower case spelling in $\log$ and $\arg$)
$$
(i^4)^{1/2}=e^{1/2 \left( \text{log}\,i^4\right)} 
=e^{1/2(\log|i^4|+i\,\text{arg}(i^4))}
=e^{1/2(\log|1|+i2k\pi))}
=e^{ik\pi}=\pm1
$$
where $k$ is an integer.
