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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!

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    $\begingroup$ Counter-example for (3): $r(z)=\sqrt z,h(z)=0$. Now $(h\circ r)(z)=0$ doesn't have branch points, though $r$ does. Less trivial counter-example (to show that $h$ doesn't have to be constant): $r(z)=\sqrt z,h(z)=z^2$. Again $(h\circ r)(z)=z$ doesn't have branch points, though $r$ does (and $r\circ h$ does). $\endgroup$
    – mr_e_man
    Nov 3, 2018 at 23:28
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    $\begingroup$ Counter-example for (4): $\sqrt{-2-1}\sqrt{-2+1}=i\sqrt3\,i=-\sqrt3\neq\sqrt3=\sqrt{(-2)^2-1}$ $\endgroup$
    – mr_e_man
    Nov 3, 2018 at 23:34
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    $\begingroup$ As for (2), the definition of elementary functions includes roots of arbitrary polynomials (not just the familiar roots of $z^n-a$), but these are "implicit functions", so I doubt that they can be represented using $\ln$. If such functions are excluded, then elementary functions are defined as being produced by addition, subtraction, multiplication, division, exponential, and logarithm. In particular, $\sqrt[n]{z}=\exp(\frac1n\ln z)$, and $\cos z=\frac12(\exp(iz)+\exp(-iz))$, and $\arccos z=-i\ln(z+\sqrt{z^2-1})$, are all elementary. I'm also not sure about the second part of (2). $\endgroup$
    – mr_e_man
    Nov 4, 2018 at 1:10
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    $\begingroup$ A branch cut is basically a step discontinuity along a curve. The logarithm, along the negative real axis, has $$\lim_{y\to0^+}\ln(x+iy)-\lim_{y\to0^-}\ln(x+iy)=2\pi i$$ and the square root has $$\lim_{y\to0^+}\sqrt{x+iy}-\lim_{y\to0^-}\sqrt{x+iy}=2i\sqrt{|x|}.$$ Two branch cuts (of functions being added together) will cancel if the step sizes are opposites. For your function $f$, the step sizes of $\ln(z-1)$ and $\ln(z+1)$ (along that line $x<-1$) add up to $4\pi i$. Half of that is $2\pi i$. So the exponent itself is discontinuous, but $e^{z+2\pi i}=e^z$, so $f$ is continuous there. $\endgroup$
    – mr_e_man
    Nov 4, 2018 at 1:46
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    $\begingroup$ 4 is true if we accept all possible values of the logarithm. The reason for the commented counterexample above is because of this. $\endgroup$
    – Brevan Ellefsen
    Nov 4, 2018 at 3:53

1 Answer 1

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(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 $\{z < -1\}$. Plot of <span class=$f_3$">

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 $\sqrt{z^2 - 1} = \sqrt{2} i$.

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  • $\begingroup$ Travis, thank you for great answer. I have a (probably trivial) question, in this answer: math.stackexchange.com/questions/1644336/… it is said that we check whether infinity is branch point by making supstitution $x=1/z$ and then checking whether zero is branch point (is this general criterion that can be used always? why does it work?). If I do that for my original example then I can conclude that infinity is not a branch point. Then, I can conclude that branch cuts cancel because if they did not then infinity would be branch point. Is this argument correct? $\endgroup$
    – Thom
    Nov 7, 2018 at 23:51
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    $\begingroup$ That method is correct---and it's correct because $z \mapsto 1 / z$ is just a conjugated reflection of the Riemann sphere, so that composing with that map merely moves branch points around but doesn't change their character. $\endgroup$
    – Travis Willse
    Nov 8, 2018 at 2:16
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    $\begingroup$ And you're welcome, I'm glad you found it helpful. $\endgroup$
    – Travis Willse
    Nov 8, 2018 at 2:17

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