# How to determine if $z+\sqrt{z-1}$ and $\frac{\sin{\sqrt{z}}}{\sqrt{z}}$ is a multi-valued function?

What I know:

I understand that for a complex function to be a multi-valued funciton they must have some branch points. I understand branch points as this definition (translated into English from my textbook):

Suppose that $$w = f(z)$$. $$z_0$$ is a branch point if and only if $$\exists r \gt 0$$, such that when $$z$$ rotates around the circle $$|z_0-z|\lt r$$, $$w$$ does not return to its original value, and that when $$\lim_{r\to 0}$$, $$w$$ still does not return to its original value.

Methods of Mathematical Physics, 3rd edition, Peking University Press, p.22

In simple compelx functions like $$f(z) = \sqrt{z-a}$$, it is straightforward to apply the above definition in the complex plane and identify that one of $$z_0$$ is $$z_0 = a$$ and thus deduce $$|w|=\sqrt{|z-a|}, \arg w = \frac{1}{2}\arg(z-a)$$ such that when $$z$$ rotates around $$a$$ for $$2\pi$$, $$w$$ only rotates by $$\pi$$, thus the multi-value property.

What I don't know:

• How to determine if a more complicated complex function is a multi-valued function?
• How to find branch points in general

Some concrete examples to what I don't know:

• How to determine if $$z+\sqrt{z-1}$$ and $$\frac{\sin{\sqrt{z}}}{\sqrt{z}}$$ are multi-valued functions?
• I don't know about a general rule. However when (square) roots are involved, look at the zeros. In the examples you gave $z=1$ for the first and $z=0$ (maybe) for the second. Sep 28, 2021 at 3:07
• I found some insights in solving questions in this answer, however I am still unsure if there is a more generic way to find branch points Sep 28, 2021 at 10:32
• for $z+\sqrt{z-1}$: let $f_1(z)=z$, then $f_1$ is obviously an analytic function (check with CR-condition $i\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$), so $f_1$ is single-valued. let $f_2(z)=z-1$, by the same reason we see that $f_2$ is single valued. let $f_3(z) = \sqrt{f_2(z)}$, then we can easily see that $f_3$ is multivalued, therefore $f(z) = f_1(z) + f_3(z)$ is multi-valued funtion. This is the generic way to determine if a complex function is multi-valued or not Sep 28, 2021 at 10:41

## 2 Answers

Answers first, $$\sin \sqrt{z}$$ is multi-valued function, but $$\frac{\sin \sqrt{z}}{\sqrt z}$$ is singled-value function.

Here's the idea:

Break them down into pieces, determine if each of them is single-valued or multipled-valued, then have them combined and see what happens. (be aware that a multi-valued function combined with a single-valued function can result in single-valued function, but also notice that combinations of single-valued functions are definitely single-valued functions)

I'll apply the above discussion to $$\frac{\sin \sqrt{z}}{\sqrt z}$$:

First we break it down into $$\sqrt{z}$$ and $$\sin z$$

### First Component: $$\sqrt{z}$$

Let $$z = re^{i(\theta + 2k\pi)}$$then $$w = \sqrt z = r^{1/2}e^{i(\theta /2 + k\pi)}, k \in Z$$

Since different $$k$$ result in different arguments, we have $$w_1 = +\sqrt{r}\\w_2 = -\sqrt{r}$$ So that we know $$\sqrt{z}$$ is multi-valued.

### Second Component: $$\sin z$$

By definition $$\sin z = \frac{e^{iz}-e^{-iz}}{2i}$$ we can easily see it is single-valued.

### Combination #1: $$\sin \sqrt z$$

Since $$f_1(z)=\sqrt z$$ maps $$z$$ into two slots, $$\sin z$$ will take the two results and map them into two other slots (recall that $$\sin z$$ is single-valued), so that $$\sin \sqrt z$$ is a single-valued function.

### Combination #2: $$\frac{\sin \sqrt z}{\sqrt z}$$

recall that $$w = \sqrt z$$ is multiple-valued $$w_1 = +\sqrt{r}\\w_2 = -\sqrt{r}$$ and that $$\sin z = \frac{e^{iz}-e^{-iz}}{2i}$$ such that $$\frac{\sin \sqrt z}{\sqrt z} = \frac{\sin w}{w} = \frac{e^{iw}-e^{-iw}}{2iw}$$ Now we plug $$w_1$$ and $$w_2$$ into the above equation, we can see that $$\frac{\sin \sqrt z}{\sqrt z} = \frac{\sin w_1}{w_1}= \frac{\sin w_2}{w_2}$$ Thus deducing the multi-valued to single-valued function.

Simple way to see that $$\frac{sin(\sqrt{z})}{\sqrt{z}}$$ is single valued. Use power series.

$$sin(y)=\sum\limits_{n=0}^\infty (-1)^n\frac{y^{2n+1}}{(2n+1)!}$$.
Then $$\frac{sin(\sqrt{z})}{\sqrt{z}}=\sum\limits_{n=0}^\infty \frac{(-z)^n}{(2n+1)!}$$ convergent for all $$z$$, without singularity.

• I am not very familiar with power series yet, but thanks for the insight! Sep 29, 2021 at 5:04