I am looking at the text by G. K. Batchelor, An Introduction to Fluid Dynamics, pg. 428-9. I am looking at the inverse mapping of $z = \zeta + \frac{\lambda^2}{\zeta}$ given by

$$\zeta=\frac{1}{2}\left(z+(z^2-4\lambda^2)^{\frac{1}{2}}\right),$$ as we wish $\zeta$ ~ $z$ as $|z|\rightarrow \infty$. Now assuming $|z|>\lambda$, this maps a circle into an ellipse.

Batchelor says ".. the value of $(z^2-4\lambda^2)^{\frac{1}{2}}$ is made unique here by specifying that there is a 'cut' in the $z$-plane at $-2\lambda \leq x \leq 2\lambda$, $y=0$ and the relevant branch is that which it is positive at $x>2\lambda$, $y=0$ (the negative branch of $(z^2-4\lambda^2)^{\frac{1}{2}}$ being that needed for the mapping of the region of the $z$-plane outside the ellipse on to the region of the $\zeta$-plane inside the circle $|\zeta|=c$)."

I have not studied branch points and branch cuts before. I think I understand that the branch points occur at $z=-2\lambda$ and $z=2\lambda$, and the branch cut that joins those is chosen. However, I do not really understand what is meant in bold. Any insight or suggested reading would be appreciated.

I have noticed when using this inverse mapping it is only valid in the positive real half-plane in the $z$-plane outside of the ellipse and I think this relates.


1 Answer 1


To simplify the matter, let's look at $\sqrt{z^2-1}$. For each $z\ne \pm 1$ there are two possible values of this square root. Worse yet, there is no way to define $\sqrt{z^2-1}$ so that it's single-valued and continuous in $\mathbb C \setminus \{-1,1\}$, because traveling around $1$ in a small loop brings you from one choice of the root to the other.

This is where the branch cut comes in. I claim that we can choose a value of $\sqrt{z^2-1}$ in $\mathbb C\setminus [-1,1]$ in a way that gives a holomorphic function. The idea is to

  1. choose the sign for $\sqrt{z_0^2-1}$ for some fixed real $z_0>1$
  2. use analytic continuation along paths connecting $z_0$ to other points in $\mathbb C\setminus [-1,1]$ (the path must lie in $\mathbb C\setminus [-1,1]$)
  3. check that analytic continuation produces consistent results.

The first step corresponds to "the relevant branch is that which is positive" -- it is natural to pick the positive sign for $\sqrt{z_0^2-1}$. Analytic continuation is possible along paths in $\mathbb C\setminus [-1,1]$ since there are no singularities in this domain. Homotopic paths produce the same continuation; this is a general fact. Non-homotopic paths ending at the same point differ by the number of times that they go around the segment $[-1,1]$. One must check that going around such a loop does not lead to a different value of $\sqrt{z^2-1}$. One way to do this is to deform the loop so that it lies in the exterior of the unit disk, where a Laurent series gives a single-valued branch of $z\sqrt{z^2-1}=z(1-1/z^2)^{-1/2}$.

Anyway, the bold text simply says that between two choices of sign, the author will use $+$ when $z$ is large and positive. The parenthetical remark says that the other choice of sign is useful for some different purpose.

I have noticed when using this inverse mapping it is only valid in the positive real half-plane

I don't see why. It should work in the complement of the cut. It's possible that the software you use to visualize the map does not understand where the branch cut is. In such cases, it may help to give it the formula written in another way, e.g., $z\sqrt{1-1/z^2}$.

  • $\begingroup$ Great explanation, you have been a great help in the two questions you've helped me with. And once again, you are absolutely right - the software did recognise the cut once I had made it clear. Thank you. :-) $\endgroup$ Dec 16, 2013 at 12:16

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