For a complex function to have a zero, both its real and imaginary parts must be zero at the same point in the complex plane. The straightforward method for identifying such points is to visualize as lines the zero values of each of these parts and see where they cross. This can unfortunately go awry in the context of functions with poles.

As a first example, consider the complex zeros of the sine function, with the real part in blue and imaginary in red:

As expected the intersections of blue and red lines for this singly-periodic function occur only along the real axis at multiples of π, as can be verified by showing such values. All of these intersections correspond to zeros.

The situation is rather different for the elliptic generalization of the sine function, the Jacobi elliptic sine. This function has zeros at the points

$2nK\left(m\right)+2kiK(1-m)$

where n and k are integers, and poles at the points

$2nK\left(m\right)+(2k+1)iK(1-m)$

The visualization of the crossing of lines of zero values of the elliptic sine is

Clearly half of the intersection points are not zeros at all, but poles of this doubly-periodic function.

This phenomenon was first noticed in visualizing the nontrivial zeros of the zeta function on part of its critical strip. It is well know that the first zero occurs around fourteen on the critical line, but this graphic appears to indicate otherwise:

This function has a single pole at $z=1$ , which again appears as an intersection of the real and imaginary zero values.

This phenomenon is thus not restricted to doubly-periodic functions, but any function with poles. That includes a function with a single algebraic pole in the same location as that of the zeta function,

$f\left(z\right)=\frac{1}{z-1}$

whose visualization is

or a more complicated product of poles,

$f\left(z\right)=\frac{1}{(z-1)(z+1)(z-i)(z+i)}=\frac{1}{z^4-1}$

whose visualization is now

In fact this behavior extends to any inverse polynomial, since they can all be factored over the complex plane. It applies as well to higher-order poles, as will be made clear shortly.

So what’s going on? Why does a pole look like a zero in these visualizations?

Begin with a simple pole at an arbitrary location:

$\begin{array}{l}f\left(z\right)=\frac{1}{z-c}=\frac{1}{x+iy-a-ib}\\ \phantom{f\left(z\right)}=\frac{x-a}{(x-a{)}^2+(y-b{)}^2}-i\frac{y-b}{(x-a{)}^2+(y-b{)}^2}\end{array}$

Denoting the individual separations from the real and imaginary parts of the pole with subscripted deltas, this can be written

$f\left(z\right)=\frac{{\Delta }_x}{{\Delta }_x^2+{\Delta }_y^2}-i\frac{{\Delta }_y}{{\Delta }_x^2+{\Delta }_y^2}$

For ${\Delta }_y$ nonzero, letting ${\Delta }_x\to 0$ produces a zero real value at the pole. Likewise for ${\Delta }_x$ nonzero, letting ${\Delta }_y\to 0$ produces a zero imaginary value at the pole. The pole only appears when ${\Delta }_x\to 0$ at the same time as ${\Delta }_y\to 0$ , that is when the approach to the pole is exactly in step in both directions.

Since numerical visualizations via marching squares work over discrete regions, the simple pole above appears exactly this way: a vertical line at the real part of the pole, and a horizontal line at the imaginary part of the pole. Expanding the region around poles for more complicated cases gives the same behavior near the pole, with the possibility of lines being interchanged by a multiplicative imaginary constant.

What about higher-order poles? The simple double pole

$f\left(z\right)=\frac{1}{(z-1{)}^2}$

has the visualization

Squaring the expression above for the simple pole in terms of deltas, the function just displayed can be represented more generally as

$f\left(z\right)=\frac{{\Delta }_x^2-{\Delta }_y^2}{({\Delta }_x^2+{\Delta }_y^2{)}^2}-i\frac{2{\Delta }_x{\Delta }_y}{({\Delta }_x^2+{\Delta }_y^2{)}^2}$

The vertical and horizontal lines at the imaginary part are straightforward from the previous discussion. The slanted lines for the real part represent ${\Delta }_x=\pm {\Delta }_y$ , and as long as the separation is not zero the pole again appears to be a zero, since the numerical evaluation occurs over a discrete region around the pole.

Similar behavior holds for all higher-order poles, which would be clear from further visualizations. Rather than display those, consider that the numerator of a higher-order pole is the expression

$({\Delta }_x-i{\Delta }_y{)}^n=\sqrt{{\Delta }_x^2+{\Delta }_y^2}exp(-in{tan}^{-1}\frac{{\Delta }_y}{{\Delta }_x})$

Setting ${\Delta }_y=c{\Delta }_x$ and expanding the exponential into trigonometric functions, the real part of the right-hand side is zero when

$cosn{tan}^{-1}c=0\to c=tan\frac{\pi }{n}(k+\frac{1}{2})$

and the imaginary part of the right-hand side is zero when

$sinn{tan}^{-1}c=0\to c=tan\frac{\pi k}{n}$

where in both cases k is an integer. For any integer n the lines ${\Delta }_y=c{\Delta }_x$ are rays emanating from the pole, which of course still appears as a zero...

It will be left for a future presentation to devise a method to make poles stand out from zeros.

Happy Groundhog Day!

Uploaded 2025.02.02 analyticphysics.com