Draw a pair of circles tangent to each other.
What's the easiest way to draw two circles like that? They sort of look like this: oo
To start off, how exactly are we ``drawing'' things here? Let's say we're drawing things parametrically in the complex plane. (This could also be correspondingly done in Cartesian coordinates.) One of the simplest things in the complex plane is the unit circle, given by either $|z|=1$, simply, or $\gamma(t)=\operatorname{cis}(t)$ for $t\in[0,2\pi]$, parametrically.
We can start making progress by shifting our one circle to make up half of our picture: $\cos t+1+i\sin t$ draws the right circle.
This is still just one circle. We want two. Arguably, taking $\gamma(t)=\cos t+1+i\sin t$ for $t\in[0,4\pi]$ draws two circles, but the two circles are the same circle drawn twice over each other. It might look like there's just one circle, but we're sort of drawing two?
How do we separate these two overlapping circles? Our desired second circle is the reflection of the first circle across the imaginary axis, i.e. $-(\cos t+1)+i\sin t$ would draw the second circle perfectly. What we want is to draw both $\cos t+1+i\sin t$ (from $0$ to $2\pi$) and $-(\cos t+1)+i\sin t$ (from $0$ to $2\pi$) in a clean, single expression.
Our slick[citation needed] solution is to take $\sigma(t)$ to be $1$ for $t\in(0,2\pi)$ and $-1$ for $t\in(2\pi,4\pi)$ (edge cases ignored). Then our two circles are given by $\sigma(t)(\cos t+1)+i\sin t$ for $t=[0,2\pi]$ and $\sigma(t)(\cos t+1)+i\sin t$ for $t=[2\pi,4\pi]$ - i.e., we can draw both with $\sigma(t)(\cos t+1)+i\sin t$ over $t=[0,4\pi]$.
(There is a rather simple choice for $\sigma$'s definition; can you find it? AnswerIt is the sign function composed with the sine function (sine's sign): e.g., take $\sigma(t):=\sin t/|\sin t|$.)
Thus, $\gamma(t)=\sigma(t)(\cos t+1)+i\sin t$ for $t\in[0,4\pi]$ does the trick.
H.A. Priestley, Ex. 4.2(iii) Define parametrically a path $\gamma$ for which $\gamma^*$ is the pair of circles $|z\pm1|=1$, the first traced clockwise and the second anticlockwise.
With two circles out of the way, though, we aren't limited to two circles at all. The trick of shift-then-reflect (strictly speaking, shift then draw the reflection) can be applied again: we shift the two circles to the right by $2$ and double it back again to get $4$ circles.
To get $4$ circles, let's take $\sigma(t)(\cos t+1)+i\sin t$ and add $2$ to it.
Now we have the right half of our desired result again, drawn by $t\in[0,4\pi]$. To get the left half drawn by $t\in[4\pi,8\pi]$, we add the negative factor to the real component again, $\sigma$ - but instead of $\sigma(t)$, we take $\sigma(t/2)$, since we want to split with the crossover at $4\pi$ instead of $2\pi$.
\[\gamma(t)=\sigma(t/2)(\sigma(t)(\cos t+1)+2)+i\sin t\qquad t\in[0,8\pi]\]
For simplicity, we can isolate the real component and form a recursion with base case $f_1(t)=\cos t$:
\[f_{j+1}(t)=\sigma(t/2^{j-1})(f_j(t)+2^{j-1}).\]
The overall parametrization is given by $f_k(t)+i\sin t$ over $t\in[0,2^k\pi]$.
We have found a not-too-complicated recursion that gives a simple formulation for arbitrary doubling of adjacent circles. However, the parameter, $t$, depends on $k$. It turns out we can readily eliminate this dependency by halving the parameter $t$ in the recursive call: \[g_{j+1}(t)=\sigma(t)(f_j(2t)+2^{j-1}),\] where $g_1(t)=\cos t$.
We parametrize this as $g_k(t)+i\sin(2^{k-1}t)$ over $t\in[0,2\pi]$. We could eliminate the additional imaginary component's dependence on $k$ at the expense of a short recursion by folding the imaginary component into the recursion, but personally I think this is the cleanest. $\Box$