Rational functions with a special symmetry

Question

Let $f\colon \mathbb{C} \to \mathbb{C}$ be a rational function that satisfies the following property: $$ f(x) = f(-1/x),\ \ \forall x\in\mathbb{C}. $$ I wonder whether functions with this property have been studied in the math (or related areas) literature and, in particular, if there are problems in which this class of functions plays a role. If so, pointers to the literature are very welcome. Thanks in advance for any comment.

Answer 1

Geometry

$g:z\mapsto -1/z$ is a reflection at the unit circle $S$, and it maps $S$ to itself. It's an isometry on the Riemann sphere $\Bbb C\cup\{\infty\}$ that maps the southern hemisphere (interior of the unit disc) to the northern hemisphere (exterior of the unit disc) and vice versa.

Thus, applying $f$ will assign values to the sphere such that the resulting sphere has similarly "colored" southern and northern hemisphere. $g$ has two fixed-points, namely $\pm i$, which means that on the Riemann sphere, $g$ represents a rotation of 180° around the axis that runs through $+i$ and $-i$.

Thus, $f$ are the rational functions on the Riemann sphere that are symmetric when rotated 180° around that axis.

Representation

Let $f(z) = c\cdot a(z) / b(z)$ be such a rational function with monic polynomials $a$ and $b$ for numerator and denominator, respectively.

$f$ can be represented by its zeros and poles, and due to the symmetry $$f(z) = f(-1/z)\tag 1$$ whenever $f$ has a zero of some order at $a_k$, then it has also a zero of that same order at $-1/a_k$, and similar for the poles of $f$, i.e. for the zeros $b_k$ of $b$. This means we can write $a$ and $b$ as:

$$\begin{align} a(z) &= \prod_{k=1}^n (z-a_k)(z+1/a_k) = \prod_{k=1}^n (z^2+{\tilde a}_k z -1) \\ b(z) &= \prod_{k=1}^m (z-b_k)(z+1/b_k) = \prod_{k=1}^m (z^2+{\tilde b}_k z -1) \\ \tag 2 \end{align}$$

with $\tilde z = 1/z-z$. The sets of zeros and poles are disjoint, i.e. $\gcd(a,b) = 1$, and terms are repeated according to their multiplicity, i.e. the degrees of $a$ and $b$ are $\partial a = 2n$ and $\partial b = 2m$, respectively.

Moreover, in order not to handle poles and zeros twice, we need some additional restriction over which zeros and poles we extend $(2)$, say

$$|a_k|\leqslant 1\ \text{ and }\ (0\leqslant \arg a_k <\pi \text{ if } |a_k| = 1) \tag 3$$ and similar for the $b_k$.

The representation $(2)$ can only handle zeros and poles that are neither at $0$ nor at $\infty$, however the behaviour at $\infty$ is already given by the degrees of $a$ and $b$, and we can handle the cases $0$ and $\infty$ by multiplying $f$ with a factor $z^j$:

$$f(z)=c\,z^j\,\frac{a(z)}{b(z)}$$

Then $f$ has a zero of order $j$ at $z=0$ and it must also have a zero of order $j$ at $\infty$, that is $j = \partial b - (j + \partial a)$ and thus $j=(\partial b -\partial a)/2 = m-n$. The same reasoning catches poles at $z=0$ so that finally

$$\begin{align} f(z) &= c\, z^{m-n}\,\frac{a(z)}{b(z)} \\ &= c\, z^{m-n}\,\frac{\prod_{k=1}^n (z^2+{\tilde a}_kz -1)}{\prod_{k=1}^m (z^2+{\tilde b}_kz -1)} \tag 4\\ &= c\, \frac{\prod_{k=1}^n (z+{\tilde a}_k -1/z)}{\prod_{k=1}^m (z+{\tilde b}_k -1/z)} \tag 5\\ \end {align}$$

The usual representation of a rational function as fraction of polynomials is given by $(4)$. Representation $(5)$ is closely related to the functional equation $(1)$.