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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?
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  • $\begingroup$ 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

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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.

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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.

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

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