TITLE: What is the difference between $x^{1/2}$ and $\sqrt{x}$?
QUESTION [5 upvotes]: 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 $\sqrt{x} \neq x^{1/2}$?

REPLY [2 votes]: 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.