Intereting Posts

Pick the highest of two (or $n$) independent uniformly distributed random numbers – average value?
Methods for efficiently factoring the cubic polynomial $x^3 + 1$
If $A$ and $B$ are positive definite, then is $B^{-1} – A^{-1}$ positive semidefinite?
counting triangles in a graph or its complement
Integral using Parseval's Theorem
Is there a proof of the irrationality of $\sqrt{2}$ that involves modular arithmetic?
using the same symbol for dependent variable and function?
Basis for a topology with a countable number of sets
If $F(a_1,\ldots,a_k)=0$ whenever $a_1,\ldots,a_k$ are integers such that $f(x)=x^k-a_1x^{k-1}-\cdots-a_k$ is irreducible, then $F\equiv0$
Integrating Reciprocals of Polynomials
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Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
Minimal polynomial of $\alpha^2$ given the minimal polynomial of $\alpha$
What is the minimum number of locks on the cabinet that would satisfy these conditions?
To what extent is it possible to generalise a natural bijection between trees and $7$-tuples of trees, suggested by divergent series?

I was trying to clarify some questions I had about elliptic integrals using

http://websites.math.leidenuniv.nl/algebra/ellcurves.pdf

There they define the map

$$\phi\colon w\mapsto \int_0^w\frac{\mathrm{d}z}{\sqrt{1-z^2}}$$

on $\mathbb{C}\setminus[-1,1]$ to get $\phi$ well-defined up to periods of the integral. The choice of the interval $[-1,1]$ is made so that $\sqrt{1-z^2}$ admits a single-valued branch.

- Derive branch cuts for $\log(\sqrt{1-z^2} + iz)$ as $(-\infty,-1)$ and $(1,\infty)$?
- How to visualize $f(x) = (-2)^x$
- Branch cut for $\sqrt{1-z^{2}}$ and Taylor's expansion!
- Contour integral of $\sqrt{z^{2}+a^{2}}$
- Understanding branch cuts by manually choosing the branch cuts of a function
- About the branch-cut in the complex logarithm

Now, I know that the principal branch of the square root $\sqrt{z}$ is discontinuous on the half-line $(-\infty,0)$, so to get a holomorphic map we restrict to $\mathbb{C}\setminus (-\infty,0]$. Substituting $1-z^2$ for $z$ we get that the appropriate branch cuts for the above mapping $\sqrt{1-z^2}$ would be $(-\infty,-1]$ and $[1,\infty)$, which is somewhat the opposite of the suggested interval $[-1,1]$.

From that I conclude that they didn’t choose the principal branch, otherwise for e.g. $z=2$ the map would be discontinuous.

My question is: Are both choices possible? Then there must be some way to choose another branch of $\sqrt{1-z^2}$. Is there a good way to see how to choose “elegant” branch cuts and the corresponding holomorphic branches?

A thought of my own: It should be possible to instead integrate on the Riemann sphere, using $\infty$ and not $0$ as a starting point. Then the two intervals would “swap roles”. But I don’t see how to formalize this.

- Integration using residues
- Physical or geometric meaning of complex derivative
- complex analysis(complex numbers) - Apollonius circle (derivation of proof)
- Sufficient conditions for an entire functions to be constant
- Entire function having simple zero at the Gaussian integers
- Can there be a point on a Riemann surface such that every rational function is ramified at this point?
- Holomorphic function with zero derivative is constant on an open connected set
- The integrals $ \int_{0}^{\infty} e^{-ax^{2}} \sin (bx^{2}) \,dx $ and $ \int_{0}^{\infty} e^{-ax^{2}} \sin \left( \frac{b}{x^{2}} \right) \, dx$
- Local and global logarithms
- Showing that $|f(z)| \leq \prod \limits_{k=1}^n \left|\frac{z-z_k}{1-\overline{z_k}z} \right|$

If we choose a branch of $\sqrt{z}$ with branch cut along the positive axis $[0,\infty)$, then we won’t need to integrate on the Riemann sphere, since $1-z^2\ge 0$ if and only if $1\ge z^2$, which happens if and only if $z\in [-1,1]$.

This is equivalent to Potato’s answer, which is a nice geometrical interpretation of the branch cut.

When taking a branch of $\sqrt z$, you can choose any ray emanating from the origin. In this case, for $\sqrt{1-z}$ and $\sqrt{1+z}$, we need to choose two rays emanating from $-1$ and $1$, and the author chooses them to be $[-1, \infty)$ and $[1,\infty)$.

This seems to rule out the entire interval $[-1, \infty)$ from being part of the domain. But it can be shown that the “jumps” in the branch cuts of the square root functions cancel on the interval $[1,\infty)$, so we get an analytic function on $\mathbb{C}-[-1,1]$.

It’s not too hard, and I invite you to try it as an exercise, that you get an analytic continuation across $[1,\infty)$.

This is how I think about it. Andrew’s answer in the comments is fantastic and probably better, though. For a complete theory of making such branch cuts, you need to learn about Riemann surfaces. If I recall correctly, Forster’s book on Riemann surfaces has a treatment of such functions, but it requires some background to access.

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- Hint on an exercise of Mathieu groups
- Prove that there is no smallest positive real number
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- How to formalize a variable-binding operator, such like $\frac{d}{dx}$?
- Lattice of continuous functions
- What is the smallest constant that has explicitly appeared in a published paper?