$c \in C \setminus R$ third unity root, $z \in C$. If $|z - 1| \le 1$ and $|z - c| \le 1$, then $|z| \le 1$

Question

Let $c \in C \setminus R$ a third root of the unity. If $a \in C$ and $|z - 1| \le 1$ and $|z - c| \le 1$, then prove $|z| \le 1$. This problem was taken from a Romanian mathematical magazine. I have solved $c ^ 3 = 1$, with $c \in C \setminus R$ and got $c = -\frac{1}{2} \pm \frac{\sqrt3}{2}i$. I tried dealing with the second inequality involving $z$ anc $c$, but I couldn't get anywhere. Any help?

Answer 1

$\newcommand\abs[1]{\lvert#1\rvert}$Fix $a_i \in \mathbb C$. You can see the following fact by drawing a picture: if $\abs{a_1} = \abs{a_2}$, then the circles of radius $\abs{a_i}$ centred at $a_i$ intersect at $0$ and $a_1 + a_2$, and the intersection of the corresponding discs is contained in the circle of radius $\abs{a_1 + a_2}$ centred at $0$.

Here's an algebraic proof, which shows that we don't even need $\abs{a_1} = \abs{a_2}$ for the second part: If $\abs{z - a_i}^2 = \abs z^2 - 2\Re(\overline{a_i}z) + \abs{a_i}^2 \le \abs{a_i}^2$, then $\abs z^2 \le 2\Re(\overline{a_i}z)$, so $2\abs z^2 \le 2\Re((\overline{a_1 + a_2})z) \le 2\abs{a_1 + a_2}\,\abs z$. That is, $\abs z \le \abs{a_1 + a_2}$ (either because $\abs z \ne 0$ and we divided, or because $\abs z = 0$ and it's obvious).

Now apply this with $a_1 = 1$ and $a_2 = c$.