31 July 2021 · Problems Maths

Complex chain (30/07/21 Classic)

Original puzzle

What is the locus of the tips of all chains that satisfy the below properties?

This is annoying to solve trigonometrically, so we turn to the complex numbers as a representation of the 2D plane. The length of the $(n+1)$th link is $f^n$, while its orientation angle relative to the first link is $nt$ for some arbitrary parameter $t\in\mathbb{R}$. So, we represent the $(n+1)$th link by a complex number $f^n e^{int}$ which describes its magnitude and orientation in the complex plane.

(Note that this complex representation means the first link is fixed to lie on the real axis between $0$ and $1$, without loss of generality. Also, the actual angle between the first two links is $|t-\pi|$ and not $t$).

Hence, the tip of the chain, $z(t)$, is the infinite sum of these complex chain links: \[\begin{align*}z(t)&=\sum^\infty_{n=0}f^n e^{int}\\&=\frac{1}{1-f e^{it}}\end{align*} \]noting the use of the convergent geometric series formula, which is valid as $|f e^{it}|=f<1$.

To determine the shape traced by $z(t)$, we intepret the equation \[z(t)=\frac{1}{1-fe^{it}}\]as a Möbius transformation of the complex number $e^{it}$ to the number $z(t)$. One property of Möbius transforms is that they map circles to either lines or circles. Indeed, the set of points $\{e^{it}\}$ is the unit circle; hence, its transform $\{z(t)\}$ must also be a circle. (Note that it cannot be a line because lines diverge to infinity on the complex plane, but we know that $z(t)$ must be finite as the infinite series above is convergent for $f<1$.)

As the shape is a circle, there must exist a centre point $c\in\mathbb{C}$ and a radius $r\geq 0$ such that $$|z(t)-c|=r$$

To determine the centre point, we use the fact that$$z(t)^*=z(-t)$$which implies the $\{z(t)\}$ circle is reflection-symmetric across the real axis, as every $z(t)$ has a complex conjugate partner on the other side of the real axis. Reflection symmetry lines for circles are diameters, so the real axis must contain a diameter. Given that $z(0),z(\pi)\in\mathbb{R}$, the line connecting $z(0)$ and $z(\pi)$ must be a diameter, and so the centre point must be their average:


To verify that $c=\frac{1}{1-f^2}$ and to find the radius, we check:


(The last line is true because $1-fe^{-it}$ and $1-fe^{it}$ are complex conjugates of each other, and hence have the same magnitude.)

So, we have $|z(t)-c|=\frac{f}{1-f^2}$, which is constant. This is the equation for a circular arc centred on $c=\frac{1}{1-f^2}$ with radius $r=\frac{f}{1-f^2}$.

Returning to the 2D geometry, we conclude that if (without loss of generality) the chain's first link is anchored the origin and oriented along the $x$-axis, then the chain's tip traces out a circle centred at $\big(\frac{1}{1-f^2},0\big)$ with radius $\frac{f}{1-f^2}$.

Below is an animation for $f=0.7$, noting the $2\pi$-periodicity of $z(t)$:


After I posted this solution on Twitter, @EricMentzell pointed out that if we assume the links cannot cross, then the chain might become halted at some point: