Understanding Root Geometry in Rational Functions

The derivatives of $1/(x^2+4x+13)$ look unremarkable at first: rational functions with growing numerators and the denominator raised to higher and higher powers. Their numerators, however, are far from random. Their zeros are evenly spaced in angle as seen from the roots of the denominator, and their size is governed by the distance to those roots. Chapter 4 of Rida Abu-Sokon's book develops this into a classification of reciprocal quadratics by root geometry. Here we derive the central formula, check it numerically, and separate the classical algebra from the book's interpretation.

The discriminant decides the regime

For $Q(x)=ax^2+bx+c$ with $a\neq 0$, completing the square gives $Q(x)=a\big[(x-p)^2-\Delta/(4a^2)\big]$ with $p=-b/(2a)$ and $\Delta=b^2-4ac$. Since $1/Q=\frac1a\cdot\frac{1}{(x-p)^2-\Delta/(4a^2)}$, it is enough to study three reduced functions:

For example, $2x^2-4x+10=2\big[(x-1)^2+4\big]$, so it falls in the first regime with $p=1$, $q=2$. The book writes the complex case as $x^2-2r\cos\theta\,x+r^2$ with roots $re^{\pm i\theta}$; in that notation $p=r\cos\theta$ and $q=r\sin\theta$.

A distance and an angle

In the complex regime, attach to each real $x$ two numbers built from the roots:

Geometrically, $R$ is the distance from $x$ to either root, and $\varphi$ is the direction of the vector from the lower root $p-iq$ to the point $x$ on the real axis. Equivalently $\cot\varphi=(x-p)/q$. As $x$ runs from $-\infty$ to $+\infty$, $\varphi$ decreases from $\pi$ to $0$ and passes $\pi/2$ exactly at $x=p$. Note that $R(x)^2=(x-p)^2+q^2$ is the reduced denominator itself.

The derivative formula

Partial fractions over the complex roots give $\frac{1}{(x-p)^2+q^2}=\frac{1}{2iq}\Big(\frac{1}{x-p-iq}-\frac{1}{x-p+iq}\Big)$. Each piece is differentiated by the classical rule $\frac{d^n}{dx^n}(x-c)^{-1}=(-1)^nn!\,(x-c)^{-(n+1)}$. Since $x-p\pm iq=Re^{\pm i\varphi}$, the difference of the two powers is $R^{-(n+1)}\big(e^{i(n+1)\varphi}-e^{-i(n+1)\varphi}\big)=2i\sin((n+1)\varphi)/R^{n+1}$. The factor $2i$ cancels and leaves

For $n=0$ it reduces correctly to $\sin\varphi/(qR)=1/R^2$, because $\sin\varphi=q/R$. We checked (1) against an independent computation, the Taylor coefficients of $1/Q$ obtained from the recursion that $Q\cdot(1/Q)=1$ imposes, for several values of $p$, $q$, $x$ and for $n$ up to $6$; the agreement is at the level of rounding error.

Two quick consistency checks are worth doing by hand. For $n=1$, $\sin2\varphi=2\sin\varphi\cos\varphi=2q(x-p)/R^2$, so (1) gives $-2(x-p)/R^4$, which is the quotient-rule derivative of $1/\big((x-p)^2+q^2\big)$. More generally, $R^{n+1}\sin((n+1)\varphi)$ is the imaginary part of $(x-p+iq)^{n+1}$, so (1) can be written without angles as

The numerator $\operatorname{Im}[(x-p+iq)^{n+1}]/q$ is a real polynomial of degree $n$ with leading coefficient $n+1$. For $n=2$ it is $3(x-p)^2-q^2$. The angular form and the polynomial form carry the same information; the angular form simply makes the zeros visible.

Angular quantization of zeros

Formula (1) locates every zero of the $n$-th derivative. Since $R>0$, the derivative vanishes exactly when $\sin((n+1)\varphi)=0$ with $0<\varphi<\pi$, that is, at $\varphi=k\pi/(n+1)$ for $k=1,\dots,n$. Translating back through $\cot\varphi=(x-p)/q$:

These are $n$ distinct real numbers, symmetric about $p$. The numerator of the $n$-th derivative, written over $\big((x-p)^2+q^2\big)^{n+1}$, is a polynomial of degree $n$, so (2) accounts for all of its zeros, and all of them are real and simple. The book calls this the angular quantization of zeros: seen from the lower root, the zeros divide the half-turn into $n+1$ equal angles.

Radial decay and angular amplification

Formula (1) separates the derivative into three factors with distinct roles. The factorial $n!$ is combinatorial growth. The radial factor $1/(qR^{n+1})$ depends only on the distance to the roots: it is largest at $x=p$, where $R=q$, and decays like $|x|^{-(n+1)}$ far away. The angular factor $\sin((n+1)\varphi)$ oscillates, and the multiplier $n+1$ means that each further derivative adds one more sign change. Together they give the envelope

with equality where $\sin((n+1)\varphi)=\pm1$. The book summarizes this as "higher derivatives = angular amplification + radial decay", with everything controlled by the root geometry. A small $q$, roots close to the real axis, makes the envelope tall and narrow near $p$. The same distance has a classical meaning: the Taylor series of $1/\big((x-p)^2+q^2\big)$ about a point $x$ has radius of convergence exactly $R(x)$, the distance from $x$ to the nearest root.

Double roots and real roots

When $\Delta=0$ the geometry collapses: $\frac{d^n}{dx^n}(x-p)^{-2}=(-1)^n(n+1)!\,(x-p)^{-(n+2)}$, which has no zeros at all. In (2), letting $q\to0$ sends every $x_k$ to $p$, which is consistent with this picture.

When $\Delta>0$ the roots $p\pm q$ are real and the trigonometric functions become hyperbolic ones. For $x-p>q$ put $R=\sqrt{(x-p)^2-q^2}$ and $\psi=\operatorname{artanh}\big(q/(x-p)\big)$, so that $x-p\mp q=Re^{\mp\psi}$. The same partial-fraction argument gives

and the even symmetry of the function about $p$ covers $x-p<-q$. Because $\sinh((n+1)\psi)>0$ there, the derivative has no zeros outside the roots. Its zeros in the complex plane solve $\big((x-p+q)/(x-p-q)\big)^{n+1}=1$, which gives $x=p+iq\cot(k\pi/(n+1))$ for $k=1,\dots,n$: the same cotangent pattern as (2), turned through a right angle onto the vertical line through $p$. The only real one is $x=p$, and only when $n$ is odd.

Classical algebra and the book's framework