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?
  • $\begingroup$ 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. $\endgroup$ Sep 28, 2021 at 3:07
  • $\begingroup$ 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 $\endgroup$
    – Ian Hsiao
    Sep 28, 2021 at 10:32
  • $\begingroup$ 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 $\endgroup$
    – Ian Hsiao
    Sep 28, 2021 at 10:41

2 Answers 2


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.

  • $\begingroup$ I am not very familiar with power series yet, but thanks for the insight! $\endgroup$
    – Ian Hsiao
    Sep 29, 2021 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.