Is the complex square root of $z^2 = \pm z$? Is $\sqrt{z^2} = \pm z$, for $z$ complex?
I think it is, since either $-z$ or $+z$ satisfies the definition
$\sqrt{z^2}= e^{\large \frac{1}{2}\log(z)^2}$
but I just wanted to make sure.  It's a bit tricky going from the positive square to the complex square root.
Thanks,

The notation $\sqrt x$ is usually avoided except when referring to the non-negative square root of a non-negative real number, because otherwise  it's ambiguous. In $C$ it is better to refer to "a square root" not "the square root".
 You will also come across the phrase "$x$ is an $n$th root of $1$" (when it is understood that $n$ is a positive integer) which means that $x^n=1$ and does not specify or restrict which of $n$ possible values $x$ might be.
