# Branch cut of $\sqrt{z^2-1}$.

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.

• What do you mean by these inequalities with complex numbers? Commented May 5, 2020 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. Commented May 5, 2020 at 23:52
• Could you please tell me where are you reading this from? Commented May 5, 2020 at 23:59
• Choose branch cuts from $-1$ to $infty$ and from $1$ to $\infty$. Commented May 6, 2020 at 0:56
• Those cuts overlap from $1$ to $\infty$. And on the overlapped part, $\sqrt{z^2-1}$ is continuous. That leaves the segment $[-1,1]$. Commented May 6, 2020 at 2:08

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 $$\sqrt{z^2-1}=-\sqrt{|z^2-1|}$$.

• 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. Commented May 6, 2020 at 4:07
• 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 Commented May 6, 2020 at 4:08
• 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 $z+1$. Commented May 6, 2020 at 4:25
• You're really quite welcome. I enjoyed helping you. And I understand how all of this can be confusing at times. Commented May 6, 2020 at 19:40
• @MeetPatel I suggest that you read the very first sentence in the posted question. Commented Mar 13 at 16:12