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,

For every nonzero complex number $w$ there are exactly two complex numbers $u_1$ and $u_2$ such that $u_1^2=w=u_2^2$.
Existence can be established directly or with the sledgehammer provided by the so-called fundamental theorem of algebra (any polynomial of positive degree with complex coefficients factors into linear factors).
Uniqueness follows from the fact that a polynomial of degree $n$ has at most $n$ distinct roots.
The polynomial you're interested in is $X^2-z^2=(X-z)(X+z)$. Since you know two of its roots, you're done: they are all the roots.
