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,
 A: 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.
A: 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.
