And what if we for example shifted one of the roots, eg $f(z)=(z+ \sqrt{z-3})^{1/3}$?

I already asked a more extensive version of this question here Branch cut/ points for square roots inside cubic roots- incorrect branching by mathematica or my mistake?,

but I am hoping that by reducing the complexity of the example and focusing on just one equation, someone could give me an understanding of how to do a complete analysis of this (and perhaps generic) multivalued function, describing its branches, and analytic structure over some cover of the complex plane.


2 Answers 2


For $~\theta \in \Bbb{R}, ~$ let $~e^{i\theta}~$ denote $~\cos(\theta) + i\sin(\theta).$

For $~n \in \Bbb{Z_{\geq 2}}, ~$ let $~S_n~$ denote the set $\{0,1,2,\cdots,n-1\}.$

For $~n \in \Bbb{Z_{\geq 2}}, ~k \in S_n,$
let $~\xi_{k,n}~$ denote exp$\displaystyle[i~(2k\pi/n)] \implies \left( ~\xi_{k,n} \right)^n = 1.$

Throughout this answer, I will be expressing complex numbers in polar coordinates. The expression $~z = re^{i\theta}~$ is to signify:

  • If $~z = 0, ~$ then $z = re^{i\theta} ~: ~r = 0, ~\theta = 0.$

  • If $~z \neq 0,~$ then $~z = re^{i\theta} ~: r \in \Bbb{R^+}, ~-\pi < \theta \leq \pi.$

Given a complex $~z = re^{i\theta}, ~n \in \Bbb{Z_{\geq 2}},$
let the function $~P(z,n)~$ represent the principal $~n$-th root of $~z$
and let it denote $~P(z,n) = r^{(1/n)} ~e^{i\theta/n}.~$

Then, when $~z_0 \neq 0,~$ the following set represents the $~n~$ distinct roots of the equation $~z^n = z_0~:$
$\{ ~P(z_0,n) \times \xi_{k,n} ~: ~k \in S_n ~\}.$

In the specific case where (for example) $~z_0 = 0,~$ and $~n = 2,~$
the set $\{ ~P(z_0,n) \times \xi_{k,n} ~: ~k \in S_n ~\}$
actually represents the two expressions
$0 \times e^{i0}, ~0 \times e^{i\pi}.~$

With the syntax in place, the problem can now be attacked.

For any complex number $~z_0:$

Express $~w = (z_0 - 3)~$ in polar coordinates as $~re^{i\theta}.$

For $~k \in S_2,~$ let $~w_k~$ denote $~P(w,2) \times \xi_{k,2}.$

That is, $~P(w,2)~$ is being used to denote the principal square root of $~w = (z_0 - 3),~$ and $~w_0, ~w_1 ~$ are being used to denote the two square roots of $~w.~$

For each $~k \in S_2,~$
let $~v_k~$ denote $~z_0 + w_k.~$

Then, for each $~k \in S_2,~$
$~P(v_k,3)~$ will denote the principal cube root of $~v_k.$

Then, the $~6~$ cube roots of

$$\left(z_0 + \sqrt{z_0 - 3} ~\right)^{(1/3)}$$

may be expressed as the following set:

$$\{ ~P(v_k,3) \times \xi_{m,3} ~: ~k \in S_2, ~m \in S_3~\}.$$

  • $\begingroup$ Thank you, that answers a part of my question. I understand that one has 6 possible roots- but what is the riemann surface which is the domain of this 6 times cover of the complex plane? where are the different branches and how should they be glued? Moreover, what if the square root was also centred around the origin instead of 3? it is still a little unclear how to analyze all of the above with the formalism you gave (although it is certainly much clearer and makes finding the roots precise) $\endgroup$
    – Noam
    Apr 1 at 14:13
  • $\begingroup$ @Noam With respect to your question about a reimann surface, my knowledge of Complex Analysis is too superficial to give a response. With respect to the other questions that you raised (e.g. what if $~z=3~$), the Math around this is very convoluted. However, as your comment indicates, except for the Reimann surface question, all of your other questions may be attacked directly by the OP (i.e. original poster), because for each such pertinent value of $~z,~$ you have an algorithm that allows you to formally (i.e. unambiguously) derive the corresponding $~(3 \times 2)~$ expressions. $\endgroup$ Apr 1 at 18:26

Try to use the more general formulation in polar form:

Let $z = re^{i(\theta+2\pi n )}$ with $z\neq 0$, $-\pi<\theta\leq \pi$ and $n\in \mathbb{Z}$

\begin{align*} z+\sqrt{z} = &re^{i(\theta+2\pi n) } + e^{\frac{1}{2}\ln(z)} \\ =& re^{i(\theta+2\pi n)}+e^{\frac{1}{2}\ln(r)+\frac{i}{2}(\theta+2\pi n)}\\ =& re^{i(\theta+2\pi n)}+r e^{\frac{i}{2}(\theta+2\pi n)}\\ =& r\left[\cos(\theta+2\pi n) + i\sin(\theta+2\pi n) + \cos\left(\frac{\theta}{2}+\pi n\right) + i\sin\left(\frac{\theta}{2}+\pi n\right) \right] \end{align*} Now note

$$|z+\sqrt{z}| = 2\left|\cos\left(\frac{\theta}{4}+\frac{\pi n}{2}\right)\right| $$


$$\arg\left(z+\sqrt{z}\right) = \frac{3\theta}{4} + \frac{3\pi n}{2} +2\pi j \quad j,n\in \mathbb{Z} $$


\begin{align*} \left(z+\sqrt{z}\right)^{\frac{1}{3}} =& \left(2\left|\cos\left(\frac{\theta}{4}+\frac{\pi n}{2}\right)\right|e^{i\left(\frac{3\theta}{4} + \frac{3\pi n}{2}+2\pi j\right)}\right)^{\frac{1}{3}}\\ =& \left(2\left|\cos\left(\frac{\theta}{4}+\frac{\pi n}{2}\right)\right|\right)^{\frac{1}{3}}e^{i\left(\frac{\theta}{4} + \frac{\pi n}{2} +\frac{2\pi j}{3} \right)} \quad n,j\in \mathbb{Z} \end{align*}

Here $n=0,1$ and $j=0,1,2$ are the roots noted in the other answer.

As an example, given that Wolfram uses the principal branch of $\ln(z)$ (ie. $n,j=0$) we have the following

$$\left(i+\sqrt{i}\right)^{\frac{1}{3}} = (2+\sqrt{2})^{\frac{1}{6}} e^{\frac{i\pi}{8}} = 1.13369733..+i0.469592... $$

which is the same result that Wolfram shows.

Edit. What are then the branch cut and the branch points?

When we define the $\ln(z)$ function over the argument $-\pi <\theta< \pi $ we say the ray $ \,\theta = \pi $ is the branch cut for the principal branch.

However, we could have defined the $\ln(z)$ over $\alpha <\theta < \alpha +2\pi$ and the branch cut would be the ray $\theta = \alpha$. In our previous example varying $n,j$ gave us a different branches for each root.

When we use the function

$$ \sqrt{z} = e^{\frac{1}{2}\ln(z)}$$

The function inherits the same branch cut used for $\ln(z)$

So, when we put $n,j=0$ in for $(z+\sqrt{z})^{\frac{1}{3}}$ the branch cut is simply the same ray we used for $\ln(z)$, ie. the ray $ \, \theta = \pi$

The branch points are those singular points where we cannot use the polar form representation, i.e. where the radicand is zero (excluding the points with polar representation that are part of a branch cut).

In our past example, the function has a branch point when

  1. the radicand of $\sqrt{z}$ is zero.
  2. the radicand of $(z+\sqrt{z})^{\frac{1}{3}} $ is zero.

The only number that satifies these conditons is $z=0$. So the branch point is simply $z=0$.

Let's see you other example $(z+\sqrt{z-3})^{\frac{1}{3}}$

Let $z=a+ib$


$$z-3 = 0 \Longleftrightarrow a+ib -3 =0 \Longleftrightarrow (a,b) = (3,0) $$

Given that there are no solution to $z+\sqrt{z-3}=0$ the only branch point is $z = 3$.

All the branch cuts and branch points are called singular points.

  • $\begingroup$ Thank you! so from this, can we systematically deduce the branch cut structure and the possible gluing one would need to make this a smooth function over some cover of the complex plane? $\endgroup$
    – Noam
    Apr 1 at 14:21
  • $\begingroup$ Hi, I made some edits to my answer. Hope this helps. $\endgroup$
    – Bertrand87
    Apr 1 at 15:51
  • $\begingroup$ Thank you, this was helpful. I am unsure about two things: are you saying we only have a branch point for the first example if both 1. and 2. are satisfied simultaneously? because 2. can also equal zero when $\sqrt{z}=-1$ correct? And similarly, for the example with z-3, don't we also have the solutions $z=\frac12 \pm i \sqrt{\frac{11}{4}}$? $\endgroup$
    – Noam
    Apr 1 at 16:10
  • $\begingroup$ These equations such as $\sqrt{z}=-1$ are not well defined precisely because you have a branch cut at $e^{\pi i}=-1$, but this is considered a branch cut not a branch point. Otherwise you would get things like $i^2=\sqrt{-1} \sqrt{-1}= \sqrt{(-1)(-1)} =1$. When dealing with multivalued functions always transform complex numbers to the most general polar form otherwise you will get in trouble. $\endgroup$
    – Bertrand87
    Apr 1 at 18:12
  • $\begingroup$ I deleted my first comment because I confused branch point with branch cut, sorry. $\endgroup$
    – Bertrand87
    Apr 1 at 18:14

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