11
$\begingroup$

For $f(z) = \sqrt{z^2+1}$, how can I find the branch points and cuts?

I took $z=re^{i\theta+2n\pi}$ and substitute in $f(z)$

$$\sqrt{r^2 e^{i(2\theta +4n\pi)}+e^{i 2k\pi}}=$$ then, I don't know how to deal with this any more
and by guessing,

I think the branch points should be $i,-i$ and cut is $[-i,i]$

$\endgroup$

1 Answer 1

19
$\begingroup$

Your solution is correct, but since you are guessing, I will explain it.

The values of $z$ that make the expression under the square root zero will be branch points; that is, $z = \pm i$ are branch points. Let $z - i = r_1e^{i\theta_1}$ and $z +i = r_2e^{i\theta_2}$. Then $f(z) = \sqrt{z^2 + 1} = \sqrt{r_1r_2}e^{i(\theta_1+\theta_2)/2}$.

  1. If we don't encircle any branch point, after one revolution, $f(z)\mapsto f(z)$.
  2. If we encircle $z=i$ but not $z = -i$, then $$ \sqrt{r_1}e^{i(\theta_1+2\pi)/2} = \sqrt{r_1}e^{i\theta_1/2}e^{\pi i} = -\sqrt{r_1}e^{i\theta_1/2} $$ Therefore, $f(z)\mapsto -f(z)$ which is multiple valued
  3. Same thing happens when we encircle $z=-i$ but not $z=i$
  4. Lets encircle both branch points. $$ \sqrt{r_1r_2}e^{i(\theta_1+\theta_2+2\pi+2\pi)/2} = \sqrt{r_1r_2}e^{i(\theta_1+\theta_2)/2}e^{2\pi i} = \sqrt{r_1r_2}e^{i(\theta_1+\theta_2)/2}\cdot 1 $$ So $f(z)\mapsto f(z)$ still single valued.

We could choose $[i, \infty)$ and $[-i, -\infty)$, but from item 4, we have seen traversing around both points returns the function to its original value. Therefore, we can choose a finite branch cut, namely, $[-i, i]$.

$\endgroup$
9
  • $\begingroup$ Your explanation is great, but I was wondering if in point 2 the $r_2$ is not missing, that is, would be the expression like $\sqrt{r_1 r_2}e^{i(\theta_1 + 2\pi)/2}$? $\endgroup$
    – David
    Commented May 20, 2018 at 21:39
  • $\begingroup$ @David in point 2 I am speaking about circling only one point so the radius is $r_1$. We get $r_1\cdot r_2$ when we circle both points. I haven't done this stuff in awhile so take this thought with a grain of salt. It could be wrong but this what I am currently thinking by looking at this. $\endgroup$
    – dustin
    Commented May 21, 2018 at 13:53
  • 1
    $\begingroup$ What does encircling a branch point mean? $\endgroup$
    – sepehr78
    Commented Apr 19, 2019 at 9:44
  • 1
    $\begingroup$ @dustin what is the range of the angles $\theta_1$ and $\theta_2$, $(-\pi,\pi]$ or $[0,2\pi)$? $\endgroup$ Commented Dec 16, 2019 at 19:27
  • 1
    $\begingroup$ @ShubhrajitBhattacharya yes, of course $\endgroup$
    – dustin
    Commented Feb 17, 2021 at 5:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .