# Branch cuts and continuity of $f(z) = \sqrt{z^2-1}$ while visualizing it

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. 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.) 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?

• 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 $\sqrt{a}$, $\sqrt{b}$, $\sqrt{ab} \ge 0$. Jun 22, 2022 at 0:20
• 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 $\pi/2 < \arg f(z) < \pi$. Jun 22, 2022 at 0:29
• @LeeMosher thanks, but I don't think this was my mistake. Jun 22, 2022 at 1:08
• @aschepler Thanks, I've reworded my question. I think I'm trying to understand the domain-colored plot better. Jun 22, 2022 at 1:09
• 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). Jun 22, 2022 at 13:31

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 $$z\in\mathbb{R}\land |z|<1$$.
• 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. Jun 28, 2022 at 6:59