TITLE: Number of Reflections in a Circle between Two Points
QUESTION [8 upvotes]: For my research I am interested in the transmission characteristics between a transmitter (Tx) and a receiver (Rx) situated in a circular room. In particular, it is important for me to know the number of paths a ray can take such that it reflects exactly once off the walls of the room. 

Since reflections occur such that the incident ray has the same angle relative to the normal as the reflected ray, I tried to use vectors to attack the problem but the math became very unwieldy.
Empirically, I have found that depending on the situation of the transmitter and receiver, there could be 2, 3, or 4 paths—no more, no less. There is an exceptional case where the transmitter and receiver are co-located at the centre, in which case there are infinitely many paths.
Can my experimental result be validated (or denied) analytically?

REPLY [13 votes]: We use complex numbers to prove that there are at most $4$ such points unless both transmitter and receiver are at the center.
Identify the circular room with the unit circle $|z|=1$ in the complex plane,
and let $r$ and $t$ be the complex numbers corresponding to the
receiver and transmitter, with $|r|<1$ and $|t|<1$.
If $z$ is a point of reflection then the condition $\theta_r = \theta_t$ 
comes down to $(z-r)(z-t)$ being a real multiple of $z^2$;
that is, to the ratio $(z-r)(z-t)/z^2$ being a real number.
Write
$$
(z-r)(z-t) / z^2 = 1 - (r+t) z^{-1} + rt z^{-2},
$$
and note that a complex number $w$ is real
if and only if it equals its own complex conjugate $\overline w$.
Since $z$ is on the unit circle, $\overline z = z^{-1}$,
so our condition is
$$
\overline{rt} z^2 - (\overline r + \overline t) z 
+ (r+t) z^{-1} - rt z^{-2} = 0.
$$
Multiplying by $z^2$ yields a polynomial of degree $4$ in $z$.
Thus there are at most $4$ solutions, even without the condition $|z|=1$,
unless the polynomial vanishes identically, in which case every $z$
is a solution.  But the polynomial vanishes identically
if and only if $r+t = rt = 0$, which is to say $r=t=0$,
so we recover the degenerate case where receiver and transmitter
are both in the center of the circular room.