$n$ insects on $|z|=1$ "occupy" a point if the product of their distances to it is at most $1$. How much of $C$ can they occupy?

The Devourers of the Unit Circle

There are $n$ tiny insects living on the unit circle $C = \{z\in \mathbb{C}, |z|=1\}$. They act as a swarm and "occupy" a point $w$ if the product of distances from $w$ to each of them is at most $1$. I.e if the insects are at points $z_j$ they occupy the set

$$\left\{w\in C : \prod_{j=1}^n |w-z_j| \leq 1 \right\}$$

How much of $C$ can they occupy at most?

Here's an interactive version of the picture below (Put the points you want to be active to the list $P_{ts}$.)

Here's what I've done so far.

Let the insects be at points $e^{ia_j}$ for $0=a_1 \leq a_2 \leq \dots \leq a_n \leq 2\pi$. Denote $f(z)=\prod_{j=1}^n |z-z_j|$. Let's study

$$g(t) = \prod_{j=1}^n |e^{it} - e^{ia_j}|^2 \\ = \prod_{j=1}^n \left|e^{-i\frac{t+a_j}{2}} \right|^2|e^{it} - e^{ia_j}|^2 \\ = \prod_{j=1}^n \left|e^{i\frac{t-a_j}{2}} - e^{-i\frac{t-a_j}{2}}\right|^2 \\ = \prod_{j=1}^n 4\sin^2 \left( \frac{t-a_j}{2} \right) \\ = 4^n \prod_{j=1}^n \sin^2 \left( \frac{t-a_j}{2} \right) $$

Now, for the occupied area to be one connected blob, in the edge case we need to find $t_1, t_2 \dots, t_{n-1}$ where $g'(t_j) = 0$ and $g(t_j) = 1$ (these are the points where the level set $f=1$ is tangent to the unit circle.

EDIT I think it's better to have the middle one (if $n$ odd) at $1$ and then others in conjugate pairs so we can collapse the pair $|(e^{it}-e^{ia})(e^{it}-e^{-ia})|^2$ into $(\cos t - \cos a)^2$ and the formulas come a bit simpler.

With the formula of WimC's answer I made this picture of how the devouring evolves as more insects arrive:

Here's the Sage-code (better version):

def getQ(n):
    a = 4^(1/n)
    return 2*a^(-n) * (1 + (1 if n%2 else -1) * chebyshev_T(n,x)(a*(x-1)+1) )

#can find the roots directly
def getRootsOfQ(n):
    a = 4^(1/n)
    #cos(theta) = a*(x-1)+1
    return [N((cos(pi*((2*m+(1-n%2))/n))-1)/a+1) for m in range(n)]

def makePic(n):
    
    x_coords = getRootsOfQ(n)
    x_coords = sorted(list(set(x_coords)))
    pts = [(x1, sqrt(1-x1^2)) for x1 in x_coords] + [(x1, -sqrt(1-x1^2)) for x1 in x_coords[:(-1 if n%2 else len(x_coords))]]
    p = lambda x,y: prod((x-b[0])^2+(y-b[1])^2 for b in pts)
    lim = 1-2*4^(-1/n)
    
    g = Graphics()
    g += circle((0,0), 1, color='black')

    var('y')
    g += contour_plot(p, (x, -1.2, 1.2), (y, -1.2, 1.2), contours=[1], fill=False)
    g += points(pts, color='blue', pointsize=40)

    return g

makePic(100).show(figsize=6)

Solution 1:

I suspect that the following construction is optimal (for a contiguous occupied arc). I did not look at your linked page so it may be the same as your work.

Assuming $P(z) = \prod(z-z_k) = z^n + \ldots + (-1)^n$ is a real polynomial (i.e. its roots are conjugacy symmetric). Then there is a real polynomial $Q(x) = (-2)^n x^n + \ldots$ such that $$Q(\frac{z+z^{-1}}2)=\lvert P(z) \rvert^2$$ for all $z\in C$. For $n$ even one can take $$Q(x) = 2\alpha^{-n}\left(1+T_n(\alpha(x-1)+1)\right)$$ where $\alpha = 4^{1/n}$ and $T_n$ is the Chebyshev polynomial of the first kind. Then $Q$ oscillates between $0$ and $1$ on the interval $[1- 2 \alpha^{-1}, 1]$. The roots of the polynomial $P$ in this case occupy an arc of angle $\beta$ where $\cos(\beta/2) = 1-2 \alpha^{-1}$.

If $n$ is odd then one can take $$Q(x) = 2 \alpha^{-n} (1 - T_n(\alpha(x-1)+1))$$ again with $\alpha=4^{1/n}$ and find the same interval $[1-2 \alpha^{-1},1]$ on which $Q$ oscillates between $0$ and $1$.