Showing that a complex number $z$ satisfies $|z|\leq 1$ given a certain condition 
If $z$ is a complex number such that $$|z-\varepsilon|\leq1\quad\mbox{ and}\quad |z-\varepsilon^{2}|\leq1$$ 
  where $\varepsilon \neq 1$ is the 3th root of unity. 
  Prove that $|z| \leq 1$.

I don't know how to start. Any ideas? 
 A: Algebraic approch. We have that
$$2=1+1\geq |z-\epsilon|^2+|z-\epsilon^2|^2=(|z|^2-2\mbox{Re}(\overline{z}\epsilon)+1)+(|z|^2-2\mbox{Re}(\overline{z}\epsilon^2)+1)$$
(recall that $|\epsilon|=1$) which implies 
$$|z|^2\leq\mbox{Re}(\overline{z}(\epsilon+\epsilon^2))=-\mbox{Re}(\overline{z})\leq |z|$$
where we used the fact that $1+\epsilon+\epsilon^2=0$ (because $\epsilon$ is a primitive root of unity).
Therefore $|z|\leq 1$.
P.S. Note that $0\leq |z|^2\leq-\mbox{Re}(\overline{z})=-\mbox{Re}(z)$ implies that $\mbox{Re}(z)\leq 0$.
A: You can do the algebra but I think there's a nice geometric argument: observe that both points $\;\epsilon,\,\epsilon^2\;$ are on the canonical unit circle, and both corresponding vectors from the origin to these two points make: $\;\epsilon\;$ makes $\;2\pi/3\;$ radians $\;=120^\circ\;$ with the positive direction of the $\;x\,-$ axis (or real axis, as you wish), while $\;\epsilon^2\;$ makes $\;4\pi/3\;$ radians $\;=240^\circ\;$ . 
Since the direct distance between these two points is $\;\sqrt3>1\;$ , the only points that could fulfil both conditions and not fulfill $\;|z|\le1\;$ are the ones on the negative axis and to the left of $\;-1\;$ , but then $\;z=x<-1\;$ , and thus
$$|z-\epsilon|=\sqrt{\left(x-\frac12\right)^2+\frac34}>\sqrt{\left(\frac12\right)^2+\frac34}=1$$
so we get a contradiction. 
A: Let $z=x+iy$ with real $x,y.$ $$ \text {We have }\quad (x+1/2)^2+(y\pm \sqrt 3 /2)^2\leq 1.$$ $$ \text {Equivalently  }\;\;  (x+1/2)^2+(\sqrt 3 /2-|y|)^2\leq (x+1/2)^2+(\sqrt 3 /2+|y|)^2\leq 1.$$ $$\text {Now we have }\quad 1\geq (x+1/2)^2+(|y|+\sqrt 3 /2)^2=$$ $$=(x+1/2)^2+|y|^2+|y|\sqrt 3+3/4 \geq (x+1/2)^2+|y|^2+3/4$$ $$ \text {which implies }\quad 1/4\geq (x+1/2)^2+|y|^2.$$ $$ \text {That is,}\quad 1/2\geq |z+1/2|.$$ $$ \text {Therefore }\quad |z|\leq |z+1/2|+|-1/2|\leq 1.$$
A: This picture may give a geometric idea for a proof:

