# Help in understanding Branch cuts and Branch points

I am trying to find out the branch cuts and branch points of the function $f(z) = \sqrt{z^2+1}$. This link gives an answer to my question? However, I have some confusions they are as follows:

1: Why Branch points of square root functions are given by the values of $z$ that make the expression under the square root zero? Does the same be applied to other multiple valued functions?

2: I need help in visualizing branch cuts of this function geometrically? In the answer (see provided the link) encircling of branch points are given. What does this mean? Can I see it geometrically through some graphs?

3: What should be the general approach to find the branch points and branch cuts of a multi-valued function?

Thank you in advance. Your answer will greatly help me to understand this concept.

• Start visualizing the different possible branches of $z^{1/2}$ then adapt it to $(z+i)^{1/2} (z-i)^{1/2} = (z^2+1)^{1/2}$. In general, if $g$ is analytic and non-zero on $U$ a simply connected open then $\frac{g'(z)}{g(z)}$ is analytic on $U$ as well as $\log g(z) =\log g(a)+ \int_a^z \frac{g'(s)}{g(s)} ds$ and $g(z)^\alpha = \exp(\alpha \log g(z))$. – reuns Aug 22 '17 at 0:08

With regard to 1) Calculus may shed a little light. If $f(z)=\sqrt{z}$, then $f'(z)=\dfrac{1}{2\sqrt{z}}$, and we see that $f'(z)$ is undefined at $z=0$ (where $f$ vanished). In terms of real variables the derivative doesn't exist at $0$, but in terms of complex variables $0$ is a non-isolated singularity, i.e. $f(z)$ is not analytic in a deleted neighborhood of $0$. The same holds true for $f(z)=\sqrt[n]{z}$.
The encircling of branch points in the linked answer you referenced is a standard approach to determining branch cuts. The issue we must avoid is a discontinuity of the function after making a complete circuit around a point. We need to ensure that traversing a closed loop produces a change in the argument of $f(z)$ which is a multiple of $2\pi$. The simplest example is $f(z)=\sqrt{z}$. A closed loop around $z=0$ produces a change in the argument of $f(z)$ by $\pi$ for a change of $2\pi$ in the argument of $z$. That is what indicates that a branch cut from $0$ to $\infty$ is required. A closed loop around $0$ cannot be allowed, because it's impossible to avoid a discontinuity of $f(z)$. (Think about the loop $e^{it}$. At $t=0, z=1, \sqrt{1}=1\text{ and }\arg f(z) = 0$, but after a circuit around $z=0$ at $t=2\pi, z=1\text{ again, but, } \sqrt{1}=-1\text{ since}\arg f(z) = \pi$.
What the linked answer shows is that a loop around either $z=i$ or $-i$ alone produces a change of $\pi$ in $\arg \sqrt{z^2+1}$ which would be a problem, but a loop enclosing both of the points produces a change of $2 \pi$ in $\arg \sqrt{z^2+1}$. A branch cut joining $z=i$ and $z=-i$ ensures that any loop enclosing one of the points must enclose the other.