Complex numbers $z$ satisfying $|z−a|+|z+a| = 2|b|\Leftrightarrow |a|\le |b|$. could you help me to prove the next statement? 
Show that there are complex numbers $z$ satisfying 
$$|z−a|+|z+a| = 2|b| $$
if and only if $|a| \le |b|$. 
I did the first implication using the triangle inequality, but I´m not able to do the second. 
Thank you in advance.
 A: The result is clear if $a=0$. Now suppose that $|a|\gt 0$.
We look for solutions of the shape $z=ta$ where $t$ is real. Then $|z-a|=|1-t||a|$ and $|z+a|=|1+t||a|$.  We want to show that  $|1-t|+|1+t|$ takes on all values $\ge 2$. This is clear by the Intermediate Value Theorem, since $|1-t|+|1+t|=2$ at $t=1$, and $|1-t|+|1+t|$ can be made arbitrarily large.
A: Notice that this set can be seen as the set of point so that the sum of distances to two fixed points ( here $a$ and $-a$ )is constant, is an ellipse, where $a, -a$ are the foci of the ellipse, and $b$ is the minor axis. Once the major axis is chosen,   by definition, the minor axis is smaller (lengthwise); if the two are equal, you have a circle. Besides, $f^2= a^2 -b^2$, so you must have $a^2 -b^2 \geq 0 $. 
A: As you see by using the Triangle Inequality, the minimum possible value of $z\mapsto |z-a|+|z+a|$ is $2|a|$, and this map is an unbounded continuous function from $\mathbb{C}\to\mathbb{R}$.  What does this imply?
P.S. You thread's title says "$|a|\geq |b|$," which is wrong.
