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I was reading something that defined the function $f(z)=\sqrt{z^2-1}$ on $\mathbb{C}\setminus [-1,1]$ where the branch cut is such that the argument of $z$ and $\sqrt{z^2-1}$ are in the same quadrant. I think I understand what this means and I think it corresponds to the usual branch of the square root.

Later, they say that $\sqrt{z^2-1}\leq 0$ for $z< -1$. I don't understand why this should be true. I have tried taking limits from above and below the imaginary axis but confused myself.

This is the way I'm understanding it: taking a point in the second quadrant slightly above the real axis, we can write $z=re^{i(\pi-\epsilon)}$ for $r>1$. Then $z^2=r^2e^{i2\pi-2\epsilon}$, i.e a complex number with argument almost $2\pi$. When you subtract one, you decrease the argument but for $\epsilon$ small it should still be nearly $2\pi$. There are two complex numbers which square to this one, one is just above the negative real axis, the other is just below the positive real axis.

To have a continuous function, we must choose the first.

curiouslearner
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    What do you mean by these inequalities with complex numbers? – Bcpicao May 05 '20 at 23:44
  • Well, then it clearly isn't true, as, no matter how you cut it, $\sqrt{z^2-1}$ should coincide with the usual square root when the argument is positive (which it does for your cutting along $[-1;1]$), meaning, $\sqrt{(-2)^2-1}=\sqrt{3}>0$. I assume there is a confusion somewhere. – Bcpicao May 05 '20 at 23:52
  • Could you please tell me where are you reading this from? – Bcpicao May 05 '20 at 23:59
  • Choose branch cuts from $-1$ to $infty$ and from $1$ to $\infty$. – Mark Viola May 06 '20 at 00:56
  • Those cuts overlap from $1$ to $\infty$. And on the overlapped part, $\sqrt{z^2-1}$ is continuous. That leaves the segment $[-1,1]$. – Mark Viola May 06 '20 at 02:08
  • Now with those cuts, you'll find that $\arg(z)$ and $arg(\sqrt{z^2-1}$ lie in the same quadrant. – Mark Viola May 06 '20 at 02:11
  • Maybe https://math.stackexchange.com/questions/289277/analytic-functions-branch helps. – Gerry Myerson May 06 '20 at 05:10
  • @GerryMyerson Thank you Gerry, I think I need a branch cut bootcamp of some sort – curiouslearner May 06 '20 at 05:34

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Let $f(z)=\sqrt{z^2-1}$ for $z\in \mathbb{C}\setminus[-1,1]$, with the branch cut on $[-1,1]$ such that $\arg(z)$ and $\arg(\sqrt{z^2-1})$ are in the same quadrant.

Branch points of $f(z)$ are at $z=-1$ and $z=1$. Corresponding branch cuts are contours that begin at $z=-1$ and $z=1$ and end at the point at infinity.

Example branch cuts include rays on the real axis from $(i)$ $z=-1$ to $z=-\infty$ and $z=1$ to $-\infty$, $(ii)$ $z=-1$ to $z=-\infty$ and $z=1$ to $\infty$, and (iii) $(i)$ $z=-1$ to $z=\infty$ and $z=1$ to $\infty$.

But the branch cuts need not be straight line paths. For example, we could choose the branch cut from $z=1$ to be hyperbolic path $\text{Im}(z)=\frac1{\text{Re}(z)}-1$ from $z=1$ to $z=i\infty$ in the first quadrant.

In terms of set equivalence (See this answer and this one for references), we can write for any value of $f(z)$ as

$$\sqrt{z^2-1}=\sqrt{z-1}\sqrt{z+1}$$

for some value of $\sqrt{z-1}$ and some value of $\sqrt{z+1}$. We choose, therefore, to cut the plane from $-1$ to $\infty$ and from $1$ to $\infty$, both along the real axis, so that

$$\begin{align} \sqrt{z^2-1}&=\sqrt{|z+1|}e^{i\arg(z+1)/2}\sqrt{|z-1|}e^{i\arg(z-1)/2}\\\\ &=\sqrt{|z^2-1|}e^{i(\arg(z+1)+\arg(z-1))/2} \end{align}$$

where $0<\arg(z+1)\le 2\pi$ and $0<\arg(z-1)\le 2\pi$. Then, $0<\arg(\sqrt{z^2-1})\le 2\pi$,

Note with these choices of branches for $\sqrt{z+1}$ and $\sqrt{z-1}$, we satisfy the requirement that $\arg(z)$ and $f(z)=\arg(\sqrt{z^2-1})$ are in the same quadrant.

Moreover, along the real axis for which $\text{Re}(z)>1$, $f(z)$ is continuous. Hence, we have now defined a function $f(z)$ that is single-valued on $\mathbb{C}\setminus[-1,1]$ and $\arg(z)$ and $\arg(f(z))$ are in the same quadrant.

Finally, note that the $\text{Re}(z)<-1$, we have $\arg(z+1)=\arg(z-1)=\pi$, $\arg(f(z))=\pi$, and $\sqrt{z^2-1}=-\sqrt{|z^2-1|}$.

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Mark Viola
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  • thank you for your answer, it is almost completely clear to me. Just a couple of questions: when you split the square root and say "as set equivalence," can you add a word or two? I am always afraid to do this with complex fractional powers. I think I know what you mean but want to make sure. – curiouslearner May 06 '20 at 04:07
  • And this is definitely pedantic, but formally should it be strict inequalities $0<\arg(z+1)<2\pi$ and same for $\arg(z-1)$ since a priori we exclude these halflines – curiouslearner May 06 '20 at 04:08
  • You're welcome. My pleasure. I've added links to a couple of references to the idea of set equivalence in the context of the complex logarithm. – Mark Viola May 06 '20 at 04:21
  • We need the non-strict inequality to define $\arg(z)$ on the entire plane. – Mark Viola May 06 '20 at 04:22
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    Thank you for the links, that is now clear. I just thought that choosing an argument is equivalent to choosing a branch of the log, so that we must exclude a ray from the origin, which here would be the positive real axis, which leads to the exclusion of $[-1,\infty)$ from z-1 and $[1,\infty)$ from $z+1$. – curiouslearner May 06 '20 at 04:25
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    You're really quite welcome. I enjoyed helping you. And I understand how all of this can be confusing at times. – Mark Viola May 06 '20 at 19:40
  • @MarkViola such that $\arg(z)$ and $\arg(\sqrt{z^2-1})$ are in the same quadrant . What is your motive to write this . please explain. – Meet Patel Mar 13 '24 at 14:43
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    @MeetPatel I suggest that you read the very first sentence in the posted question. – Mark Viola Mar 13 '24 at 16:12