Here is a line in a proof in a complex analysis text:
$\sqrt{1-z^2}=\sqrt{1-z}\sqrt{1+z}$
I know you can't do this in general, but when can you do it? Here is what I tried:
$\sqrt{1-z^2}=e^{1/2\log(1-z^2)}=e^{1/2\log((1-z)(1+z))}$
which, at this point, I would like to say is: $e^{1/2(\log(1-z)+\log(1+z))}$? But I don't believe this can necessarily be done either.
And so my question of when it's okay seems like it will depend on what branch/argument I give log.
If I choose the branch of log to be $(-\infty,0]$ then $\sqrt{1-z^2}$ only has issues on $[1,\infty)$ and $(-\infty,-1]$. So if I was on a domain that avoided these intervals, say the open unit disc...would I be able to do what I want with spitting the square root? My reasoning is this: If $arg(z) \in (0,\pi)$, then $arg(-z) \in (-\pi,0)$, and thus $arg(1+z) \in (0,\pi)$ and $arg(1-z) \in (-\pi,0)$. Thus the sum of the two arguments individually is in $(-\pi,\pi)$. Since $arg(1-z^2) \in (-\pi,\pi)$, this value must equal the sum the two individually. (Since it is well known the identity works mod $2\pi$).
This is a good question because it raises an issue where it is easy to get oneself into trouble. (Just by way of example, I recently helped my wife with an algebra assignment she got as a returning college student. Despite having been a theoretical research physicist for the last 12 years, it still took me several minutes to deduce and explain why doing the same problem twice using two different but, what I originally thought to be equivalent, orders of operation was giving two differing results. It turned out to be the two-valued nature of the square root that your question is getting at, which can be tough to see sometimes in cases involving complex numbers.)
First off, and I don't mean to sound condescending at all when I say this, I think that restating the problem in logarithmic form is unnecessary and obscures the real issue here, which I will now attempt to clarify. Clearly, the identity $(1-z^2) \equiv (1-z)(1+z)$ is true for all $z$, so the problem arises with taking the square root. To be precise, the problem is that the symbol $\sqrt{\;\;}$ is meant to represent the $principle$ square root, whereas we know that every complex number actually has two second-roots in the more general sense of numbers that square to it, and it is the throwing out of these extra possibilities that causes your identity to fail in certain circumstances. Since the two second-roots of any c-number just differ by an overall sign, the problem can be corrected by adding the $\pm$ symbol to the right-hand side. That is, $\sqrt{(1-z^2)} \equiv \pm\sqrt{(1-z)}\sqrt{(1+z)}$ is always true for $either$ the $+$ or the $-$ sign. For example, $1=(-1)^2$ clearly does not imply that $\sqrt{1}=\sqrt{-1}\sqrt{-1}=i\cdot i=-1$, but adding an overall $-$ sign corrects the problem. Put another way, the original identity fails for exactly the same reason that $x^2 = 4$ does not imply that $x=\sqrt{4}=2$. The other possibility, that $x=-2$, must be allowed for.
Now that the source of the difficulty has been thoroughly identified, we can go on to try and answer in a bit more detail when the original identity is true as given $without$ the need to throw in an extra overall minus sign. The principal square root represented by $\sqrt{\; \;}$, is defined to have a branch cut along the negative real axis. So, the identity works, for one, whenever all three of $(1-z^2)$, $(1-z)$, and $(1+z)$ have imaginary parts with the same sign. Since $(1+z)$ and $(1-z)$ have opposite imaginary parts, this possibility is only realized when $z$ is real. The other possibility is that two of their real parts have the same sign and the other one is opposite. I won't bother to figure out for you which values of $z$ that's true for (answerer's perogative ;)), but they shouldn't be too hard to deduce yourself going case-by-case.
Incidentally, it would also be worth checking whether the proof you are following breaks down when an extra minus sign is thrown in, or if maybe it doesn't matter. Hope this helps! :)