1
$\begingroup$

The branch cuts of $f(z) = \sqrt{z^2-1}$ have been discussed on this site before. However, there are certain things I still do not understand. I'm following the discussion in Morse and Feshbach's book on mathematical physics and the figure below is from Section 4.4 of the book.

complex plane

Define $g(z) = (z-1)(z+1)$ so that $f(z) = \sqrt{g(z)}$. Writing $z-1 = r_-e^{i\tau_-}$ and $z+1 = r_+e^{i\tau_+}$, we have $g(z) = r_-r_+e^{i(\tau_-+\tau_+)}$ and $f(z) = \sqrt{r_-r_+}e^{i(\tau_-+\tau_+)/2}$. Thus, $\arg(g) = \tau_- + \tau_+$ and $\arg(f) = \arg(g)/2$. The chosen branch cut is the interval $[-1,1]$ on the real axis.

Clearly, $f(z)$ is discontinuous across the branch cut: for a point in between $-1$ and $1$ and slightly above the real axis, $\tau_- \approx \pi$ and $\tau_+ \approx 0$, making $\arg(f) \approx \pi/2$. However, on bringing this point just below the real axis, we still have $\tau_+ \approx 0 $ but $\tau_- \approx -\pi$, making $\arg(f) \approx -\pi/2$. Thus, moving across the branch cut produces a discontinuity in $f(z)$.

However, plotting the function suggests that a discontinuity also exists across the entire imaginary axis. (Mathematica also produces a similar plot.)

complex plot

I'm trying to understand this plot. On the positive imaginary axis, $\tau_-+\tau_+ = \pi$, giving $\arg(g) = \pi$ and $\arg(f)=\pi/2$. For points to the right of the positive imaginary axis (1st quadrant), since $\tau_-+\tau_+ < \pi$, we have $\arg(g) < \pi$ and $\arg(f) <\pi/2$. For points to the left of the positive imaginary axis (2nd quadrant), since $\tau_-+\tau_+ > \pi$, $\arg(g)$ becomes negative and close to $-\pi$, assuming the convention $-\pi < \arg(g) \leq \pi$. Now, since $g(z)$ is "fed" to the square root function to produce $f(z)$, I'm assuming that the software takes half this value to produce an $\arg(f)$ that is close to $-\pi/2$. This is what I think is causing a discontinuity in the visual representation of the function.

However, if we choose the convention $0 \leq \arg(g) < 2\pi$, or take $\arg(f)$ to be equal to $\arg(g)/2 = (\tau_-+\tau_+)/2$, without first finding the equivalent to $\arg(g)$ in the interval $(-\pi, \pi]$, the discontinuity in $\arg(f)$ (across the imaginary axis) seems to disappear.

So my question is two fold: (i) is the plotting software producing a misleading plot? (ii) isn't $f$ continuous over the imaginary axis?

$\endgroup$
5
  • $\begingroup$ You should beware of using a formula like $\sqrt{ab} = \sqrt{a} \sqrt{b}$ that is only well-defined and valid on a restricted domain of input values, in this case $a \ge 0$ and $b \ge 0$ with output values $\sqrt{a}$, $\sqrt{b}$, $\sqrt{ab} \ge 0$. $\endgroup$
    – Lee Mosher
    Commented Jun 22, 2022 at 0:20
  • $\begingroup$ Why do you say $\arg f$ is close to $-\pi/2$ near the left side of the positive imaginary axis? In the second quadrant, $\pi < \tau_{-} + \tau_{+} < 2\pi$ so $\pi/2 < \arg f(z) < \pi$. $\endgroup$
    – aschepler
    Commented Jun 22, 2022 at 0:29
  • $\begingroup$ @LeeMosher thanks, but I don't think this was my mistake. $\endgroup$
    – B215826
    Commented Jun 22, 2022 at 1:08
  • $\begingroup$ @aschepler Thanks, I've reworded my question. I think I'm trying to understand the domain-colored plot better. $\endgroup$
    – B215826
    Commented Jun 22, 2022 at 1:09
  • 1
    $\begingroup$ If you use the primary branch cut of $\sqrt{}$ in the form $f(z) = \sqrt{z^2+1}$, and the plotting programs probably assume that, then that function is discontinuous along the imaginary axis. The form $f(z)= \sqrt{r_{-} r_{+}} e^{i(\tau_{-}+\tau_{+})/2}$ is continuous along the imaginary axis (except the origin). $\endgroup$
    – aschepler
    Commented Jun 22, 2022 at 13:31

1 Answer 1

2
$\begingroup$

Mathematica implementation of the $\arg(s)$ function has a branch cut along the negative real axis (see Sets of discontinuity) and $s=z^2-1$ is a negative real number along the entire imaginary axis since for $z=i t$ where $t\in\mathbb{R}$ then $s=z^2-1=(i t)^2-1=-t^2-1$ is always a negative real number. Therefore Mathematica's evaluation of $\arg(z^2-1)$ has a branch cut along the entire imaginary axis.

But Mathematica's evaluation of $\arg(z^2-1)$ also has a branch cut along the real axis for $z\in\mathbb{R}\land |z|<1$ since $s=z^2-1$ is also a negative real number for $z\in\mathbb{R}\land |z|<1$.

$\endgroup$
1
  • $\begingroup$ Thanks. My question was regarding $\sqrt{s}$ not $\log{s}$, but your answer applies to $\sqrt{s}$ as well. If you could edit your answer, I'll accept it. $\endgroup$
    – B215826
    Commented Jun 28, 2022 at 6:59

You must log in to answer this question.

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