Elementary questions about branch points and branch cuts Consider the following complex function: $$f(z) = \sqrt{z-1} \cdot  \sqrt{z+1}. $$
It is posible to write it as: $$f(z) = e^{\frac{1}{2}(\ln(z-1)+\ln(z+1))}. $$
(1) From $\ln(z-1)$ I get branching points $z=1$ and $z=\infty$. And position of branch cut (I use convention $\varphi \in (-\pi,\pi]$ ) is $y=0, x<1$.
(2) From $\ln(z+1)$ I get branching points $z=-1$ and $z=\infty$. And position of branch cut (I use convention $\varphi \in (-\pi,\pi]$ ) is $y=0, x<-1$.
Professor said that we have overlapping branch cuts for $x<-1$ and that they cancel each out. And therefore that $z=\infty$ is not a branch point (because branch points are points where branch cuts begin/end).
I have several questions.
(1) Do overlapping branch cuts always cancel each other? If not, what would be a counterexample? If they do, why? What does it even mean that branch cuts cancel each other? If they do cancel sometimes and sometimes they do not, does there exist a criterion by which I can tell whether they will cancel? If there is such a criterion, what is it? Why does it hold?
(2) I succeeded to write $f(z)$ in terms of logarithms. I think I would be able to do that for any elementary function (is that true?). In other words, all branch points of elementary functions are "produced" by logarithm somehow (is that true?). Does there exist a set of functions (which is not equal to the set of all functions that have branch points) that somehow "produce" branch points for all other functions which have branch points, just as logarithm produces branch points for all elementary functions? In other words, is there a set of functions with branch points which somehow "span" all functions with branch points? I am not sure this question makes any sense.
(3) Complex function $g$ is a composition of some complex functions $h$ and $r$: $$ g(z) = (h \circ r)(z).$$ If function $r$ has some branch points, why are those also branch points for $g$ (how to prove that)? If that is not even true, what would be counterexample and what would be conditions on functions $h$ and $r$ so that statement would be true?
(4) Is $f(z) = \sqrt{z-1} \cdot  \sqrt{z+1} = \sqrt{z^2-1}?$
Thank you for any help!
 A: (1)

To say that the branches of the factors cancel is to say that $f$ is still analytic along the overlap of the branches (excluding the endpoint(s) of the overlap).

Branches do not always cancel where they overlap: For example, we can generalize $f$ to

$$f_k(z) := \sqrt[k]{z - 1} \cdot \sqrt[k]{z + 1}$$

for positive integers $k$. The locations of the branch cuts of the two radical expressions (making the same choice of logarithm branch as in the question) are the same as for $f$, but the branch cuts do not "cancel" on their overlap for $k \geq 3$. For $\rho > 1$ and small $\epsilon$, $$f_k(-\rho e^{(\pi \mp \epsilon) i}) = e^{\pm 2 \pi i / k} \sqrt[k]{\rho^2 - 1} + O(\epsilon) .$$ In particular, $f_k$ is only continuous at $-\rho$ for $k = 1, 2$.
Plot of $f_3$; note the discontinuity in the argument along $\{z < -1\}$.
$f_3$">

Roughly speaking, checking continuity at an arbitrary point of the cut (excluding the branch point) is generally sufficient to establish (non)cancellation: Analyticity of the function then follows from the Inverse Function Theorem.

(2) I'm not sure how to make your question precise.
(3) This is certainly not true in general. For example, if $h$ is the zero function, so is $g = h \circ r$, but this function has no branch points. In general, the only way that $h \circ r$ can avoid inheriting a branch cut from $r$ is, roughly speaking, for $h$ to map the distant points to which $r$ maps nearby points on opposite sides of the branch cuts back to nearby points. So, continuing our example from before, taking $r = f_k$ and $h(z) := z^k$, we have $$g(-\rho e^{(\pi \mp \epsilon)}) = h\left(e^{\pm 2 \pi i / k} \sqrt[k]{\rho^2 - 1} + O(\epsilon)\right) = \left(e^{\pm 2 \pi i / k} \sqrt[k]{\rho^2 - 1} + O(\epsilon)\right)^k = \rho^2 - 1 + O(\epsilon) ,$$ so $g$ is continuous on $\{z < -1\}$, and a similar argument shows that $g$ is continuous on $\{-1 \leq z \leq 1\}$, too; in fact, $g(z) = z^2 - 1$.
(4) No. For $z = -i$, for example, $\sqrt{z - 1} \sqrt{z + 1} = -\sqrt{2} i$, but $\sqrt{z^2 - 1} = \sqrt{2} i$.
