The complex point at infinity is a rather strange duck. It completes the complex plane by adding a single “point” that circles the entire plane. As such it retains a complex phase:

$z_{\infty }=\infty ·e^{i\phi }$

This is in contrast to the real points at infinity, which are the two distinct values $\pm \infty$ , a much simpler concept. The complex version can benefit from some visualizations, especially in the context of branch cuts and the points at which they occur.

First look at the simple complex function

$f\left(z\right)=z=x+iy$

which has the straightforward visualization

The coloration here and in what follows is by complex argument. Every circular contour here about the origin is thus identical in color due the to uniform distribution of argument.

As such a contour expands in radius it approaches the point at infinity, which is always outside of the contour. One can visualize a relationship between the two by introducing an asymptotic variable:

$w=\frac{1}{z}=\frac{1}{r{e}^{i\phi }}=\frac{1}{r}{e}^{-i\phi }$

In this new variable the point at infinity becomes the origin, and any finite contour is reduced in size with negative argument. If the point at infinity is initially represented by a black circle around the contour, the relationship is this:

Any finite contour thus circles the point at infinity in the direction opposite to how it circles the origin. This interactive graphic is a nice way of seeing how the “point” at infinity is mapped to a truly single point at the origin.

Now consider the square root function

$f\left(z\right)=\pm \sqrt{z}$

This function has a branch cut that on the principal branch extends from the origin to negative real infinity. Here is how that looks:

This function has two values at all finite points other than the origin, where it has the single value of zero. It also has a single value at the point at infinity, which can be seen by again introducing an asymptotic variable:

$\sqrt{w}=\frac{1}{\sqrt{z}}=\frac{1}{\sqrt{r{e}^{i\phi }}}=\frac{1}{\sqrt{r}}{e}^{-i\phi /2}$

As $z\to \infty$ there is only one value for the inverse of the function, implying only one value of the direct function as well.

A branch cut means there are different values for the function on either side of the cut. For a finite contour around the origin, this appears as a discontinuity in coloring by argument. This can be seen nicely by repeating a previous interactive graphic with the complex argument reduced by two due to the square root:

In terms of the inverse variable the contour again circles the point at infinity in the opposite direction, but the discontinuity in coloration remains. This is the sense in which this function has a branch point at infinity.

Note that the location of the discontinuity in coloration can be altered by how one assembles parts of the branches of the complex function. For example, combining half planes on either side of the branch cut in order to provide continuity across the negative real axis moves the branch cut to the positive real axis:

In the context of the interactive graphic with the explicit point at infinity, this change would move the discontinuity in coloration to the positive real axis as well. Note that the values for the “sheet” are retained here for convenience.

Now consider a square root function with two branch points at finite values,

$f\left(z\right)=\pm \sqrt{(z-1)(z+1)}$

whose straightforward visualization is

This function appears at first sight to have branch cuts from the origin to both positive and negative imaginary infinity in addition to the cut connecting the two branch points. This initial impression can be clarified by using a method employed previously: combining half planes from different sheets. In this case one does so along the imaginary axis, giving

with the values for the “sheet” retained again for convenience. This presentation of the function makes sense since

$f\left(z\right)\to \pm z,z\to \infty$

so that far away from the cut between branch points this function behaves like the simple linear at the outset. That means a circular contour approaching infinity will have no discontinuity in coloration, so that this function has no branch points at infinity.

One is then naturally led to consider a square root function with three branch points at finite values:

$f\left(z\right)=\pm \sqrt{z(z-1)(z+1)}$

The visualization of this function again appears to have branch cuts that do not connect branch points,

but in this case they do not appear along an axis. The location of these apparent branch cuts can be determined by first separating the real and imaginary parts of the function:

$\begin{array}{c}\pm \sqrt{z(z-1)(z+1)}=\pm \sqrt{z(z^2-1)}=R+iI\\ (x+iy)(x^2-y^2-1+2ixy)=R^2-I^2+2iRI\\ x(x^2-3y^2-1)+iy(3x^2-y^2-1)=R^2-I^2+2iRI\end{array}$

From the last visualization the location in question clearly occurs when $R=0$ . Putting this into the right-hand side and equating imaginary parts gives

$y=03x^2-y^2-1=0$

This first case holds for actual branch cuts along the real axis. Using the second to select portions of the two sheets in the right-hand half plane produces the visualization

with the values for the “sheet” retained again for convenience. One can now clearly see the actual branch cuts: one between the origin and positive one, and the other extending from negative one to negative real infinity. The latter means that function does have a branch point at infinity, just like the first square root function above.

One can now begin to see a pattern for when there is or is not a branch point at infinity for square root functions. Starting with a general form and allowing the independent variable to go to infinity,

$\underset{z\to \infty }{lim}\sqrt{\prod _{k=1}^n(z-c_k)}=\underset{z\to \infty }{lim}{z}^{n/2}\sqrt{\prod _{k=1}^n(1-\frac{c_k}{z})}=e^{in\phi /2}\underset{r\to \infty }{lim}{r}^{n/2}$

there is only a change in the function traversing an entire circular contour at infinity when n is odd. That is, for a square root function with an even number of branch points, the cuts connect the branch points in pairs. When there are an odd number of branch points, one cut must extend from one of the branch points to infinity. A square root function with an odd number of branch points always has a branch point at infinity.

For more complicated functions with branch points, the same general principal will hold. It all depends on whether traversing a contour at infinity returns to the same value, or whether the cumulative contribution from all branch cuts is different from unity.

Uploaded 2025.01.31 analyticphysics.com