Branch cut for $\sqrt{1-z^{2}}$ and Taylor's expansion!

Question

I'm working in a problem that involves the equation $$ w(z)=\sqrt{1-z^{2}} \,\, . $$

I already know that there're two branch points in this equation, namely $\pm 1$, so there's a Riemann surface covering the domain of the function where the branch cut is from the $-1$ to $1$, as shown in the figure below.

My purpose is make an expansion through Maclaurin serie around the $0$ point, but I don't know if it's a suitable point to expand that function, I mean, If there's some kind of problems or inconsistency in expanding around that point. if that's not a suitable point, how could I deform the branch cut line to make this a suitable point for expansion?

Greetings!

Answer 1

What are we doing?

Branching is about multivalued functions, so it's worth stating more explicitly what we're doing.

We can take the map $z\mapsto z^{1/3}$ and expand it about a point $z_0 = r~e^{i\theta}$ to find that it has an analytic continuation which converges on the disc $|z - z_0| < r.$ We could then take a step of size, say, $r (\phi - 1)$ (where $\phi = (1+\sqrt{5})/2$ is the golden ratio) around the circle with center $0,$ to a point $z_1$ in this disc. And then we can repeat that; the general pattern is very clear: we find those derivatives, analytically continue the function about $z_n$, get another expansion valid for $|z - z_n| < r$, and take a step of length $r (\phi-1)$ to $z_{n+1} = e^{2\pi i/10} z_n$. What happens with the sequence of values $f_n$ created by this procedure?

Well, in 10 steps of this procedure we're "nominally" back where we started, in the sense that $z_{10} = r e^{i(\theta + 2\pi)}$ which is morally the same as $z_0 = r e^{i\theta}.$ But: our process of doing analytic continuations has not yielded $f_{10} = f_0.$ If we do 10 more steps, we still don't have $f_{20} = f_0.$ Only after doing 30 steps do we find ourselves both in the position that $z_0 = z_{30}$ and that $f_0 = f_{30}.$

Branching comes when we realize that complex mathematics is so simple and beautiful that we should listen to it and take it seriously when it does this. It is pushing back on us, telling us that the proper domain for $z\mapsto z^{1/3}$ is not $\mathbb C$ but is instead $\mathbb C + \mathbb C + \mathbb C.$ If you want to be rigorous we could more formally say $\{1,2,3\}\times(\mathbb C - \{0\}) + \{(0,0)\}$ where the first number is a "branch sheet index" and the product is the Cartesian product, but you're free to also think "3 copies of the complex plane knit together in the right way." It's a structure that we could make if we had a 4D Euclidean space or a 3D space with a repeating boundary condition in one dimension, with tape and sheets of paper: take 3 sheets and a pair of scissors, stack the papers $A,B,C$ and then cut them identically, then tape one edge of A to the mirroring edge of B, the leftover edge of B to the mirroring edge of C, and the leftover edges of A and B. On this geometric entity, the map $z\mapsto z^{1/3}$ becomes defined and analytic at every point which is not $0$; our difficulty is just that we took this function's "natural domain" and tried to compress it down to $\mathbb C$ by flattening the sheets of paper into each other, at which point the function became "multivalued."

One easy way to think about branch points and branch cuts, therefore, is to look at a multivalued function with $n$ branch points $b_n$ by describing, instead of $z$, the $n$ numbers $c_n = z - b_n = r_n e^{i\theta_n}$ -- but without assuming that $e^{2\pi i}$ will take you back to the same $c_n$ (as it may transition to a different branch sheet). We call these "phasors" since we're pretty much just interested in what their phases $\theta_n$ do to the resulting expression.

So then we make a "tour": we vary $z$ around the complex plane while we track the different phasors, which can be visualized straightforwardly as arrows pointing from the different branch points to $z$. We complete the tour by coming back to the "same place" on the complex plane. Then, whenever any phasor rotates by $2\pi, 4\pi, 6\pi,\dots$ we know that some sort of new branch sheet is needed unless the function overall returns to its original values for that distribution of phasor-rotations.

In the case of $z\mapsto z^{1/3}$ we see that there is one phasor and as we make a tour around the point $0$ we come to 3 different sheets before the $6\pi$ rotation causes the function to repeat itself. In the case of $z\mapsto z^{1/3} + (z-1)^{1/2}$ we need 6 different sheets, as tours which only go around $1$ do not cycle the phase of the phasor from $0$ at all (it just wobbles a bit around $\theta_0 = 0$) and repeat after 2 loops around, while tours which go around $0$ do not cycle the phase of the phasor from $1$ at all, and for no values do these start to spontaneously coincide. But for $z\mapsto z^{1/4} \times (z-1)^{1/2}$ we don't need the full 8 sheets (though we could certainly do that!) because we can get away with only 4, with the phasor from $1$ bumping our branch sheet index by 2 while the phasor from $0$ only bumps it by 1, both modulo 4: the resulting $-i^2 = 1$ is fine.

The "cuts" come about, again, when you take some scissors to the natural stack of the domains of the multivalued functions with an attempt to separate out a single sheet, which will have these tears in it from the scissors. When you tape those tears together and call it $\mathbb C,$ then of course we truly know that you cannot cross that cut without either choosing analyticity (in which case you go to a different branch sheet) or single-valuedness (in which case the Cauchy-Riemann conditions fail along the tear).

How can we do it in your problem's case?

So your branching geometry looks like this: we can factor the term inside the square root as $\sqrt{(1 + z)(1 - z)}$ and then for every $z$ we have two complex numbers $\xi = 1 + z = \alpha e^{i\phi}, \; \zeta = 1 - z = \beta e^{i\psi}.$

The square root then gives us $\sqrt{\alpha\beta}~e^{i(\phi + \psi)/2}.$ So there are two ways to look at this: one is that the valid domain of each is $0\le \phi,\psi \lt 4\pi$ before we get into periodicity ($2\times 2 = 4$ branch sheets); but it turns out that the function has the same values across 2 of those sheets and so you can reduce it down to 2. Let's call one of these the "main" branch and the other one the "off" branch.

Now we need to choose arbitrary cuts where we travel from one branch sheet to another. Again looking at the form $\sqrt{\alpha\beta}~e^{i(\phi + \psi)/2}$ we see that looping around both $\pm 1$ will cycle the phase by $2\pi$ and bring us back to the "main" branch, but looping around either one individually without the other will cycle the phase by only $\pi$ and will bring us to the "off" branch.

There are 2 choices of cuts that work for this. One choice, the "obvious" one, is to jump to the "off" branch whenever we cross the line $z = r + 0 i$ for $|r| \le 1.$ This makes what you're trying to do impossible. But a less-obvious one is to change branches whenever we cross the line $z = r + 0 i$ for $|r| \ge 1.$ This has the same properties: that closed loops around either single point will flip branches, but closed loops around both points will flip branches twice so that you're back on the original one. (If it helps, when you look at this on the sphere $\mathbb C \cup \{\infty\}$ they are basically the same, we've just chosen a branch cut that goes all the way through the point at infinity.)

Therefore yes, there are two valid Maclaurin expansions around $z=0$ with radius of convergence $1$ and they are defined on branch sheets which are not cut on that unit circle; if you try to cut them on that unit circle then you'll just have one of these Maclaurin expansions on one side of the cut and the other on the other side of that cut.

(However this is not so interesting because you can pretty much always cut from infinity into the branch points to create a single branch sheet -- the criterion for not leaving the branch sheet being "don't cycle any of the phasors by $2\pi!$". No, the more interesting thing is that you can choose the branch cut to lie between $-1$ and $1$, which means that there is a valid Laurent series about $z = 0$ with inner radius of non-convergence $1$ -- because you can also choose the branch cut in this case to lie entirely within the unit circle, this Laurent series is, in principle, possible. It is given by $\sqrt{-z^2} \sqrt{1 - z^{-2}},$ so take the same series, flip all of the exponents from positive to negative, and then multiply it by $\pm i z$ depending on which branch you're on.)

Answer 2

Your idea works fine, since $w$ can be defined as a holomorphic function in the unit disk (for example).

Use that fact that $$ \sqrt{1-x} = \sum_{k=0}^{\infty} (-1)^k\binom{1/2}{k}x^k=1+\sum_{k=1}^{\infty}\frac{(-1)^k(1/2)(1/2-1)\cdots (1/2-k+1)x^k}{k!} \\ =1+\sum_{k=1}^{\infty}\frac{(-1)^k(-1)(-3)\cdots (-2k+3)x^k}{2^k k!} \\ =1-\sum_{k=1}^{\infty}\frac{1\cdot 3 \cdots (2k-3)x^k}{2^k k!}=1-\sum_{k=1}^{\infty}\frac{(2k)!\,x^k}{4^{k}(2k-1) (k!)^2} \\ =1-\sum_{k=1}^\infty \frac{1}{2k-1}\binom{2k}{k}\left(\frac{x}{4}\right)^{\!k}, $$ and its radius of convergence is equal to 1.

Then $$ w(z)=\pm\sum_{n=0}^\infty\frac{(2n)!\,z^{2n}}{4^n(2n-1)(n!)^2}, $$ defines two holomorphic functions in the unit disk, satisfying $w^2(z)=1-z^2$.