$|z-a|+|z+a|=2|c|$ iff $|a| \leq |c|$ Full question:  let $a$ and $c$ be complex numbers, prove that there exists a complex number $z$ such that $|z-a|+|z+a|=2|c|$ iff $|a| \leq |c|$
so far I have this:
($\Longrightarrow$)
$$
|z-a| \leq |z|+|a|
$$
$$
|z+a| \leq |z|+|a|
$$
$\Longrightarrow$ $2|z|+2|a| \geq 2|c|$
I am stuck from here, and figure that solving this implication will help me figure out the other implication.  Help please!
 A: For the first implication, we can use the triangle inequality.
\begin{align}
2\lvert c\rvert & = \lvert z+a\rvert + \lvert z-a\rvert\\
&\geq\lvert z-a - (z+a)\rvert\\
&= 2\lvert a\rvert
\end{align}
Therefore, $\lvert c\rvert\geq\lvert a\rvert$.
For the second implication, if $a=0$, the result is true. For $a\neq 0$, let $z=\frac{a}{\lvert a\rvert}\lvert c\rvert$
\begin{align}
2\lvert c\rvert &=\lvert a\rvert(\lvert c\rvert/\lvert a\rvert - 1)+\lvert a\rvert(\lvert c\rvert/\lvert a\rvert + 1)\\
&=\lvert z - a\rvert + \lvert z+a\rvert
\end{align}
Now
\begin{align}
2\lvert c\rvert &= \lvert z-a\rvert + \lvert z+a\rvert\\
4\lvert c\rvert^2 &= (\lvert z-a\rvert + \lvert z+a\rvert)^2\\
&\leq 2(\lvert z\rvert^2+\lvert a\rvert^2)\\
&\leq 4(\lvert z\rvert^2+\lvert a\rvert^2)\\
\sqrt{\lvert c\rvert^2-\lvert a\rvert^2}&\leq \lvert z\rvert
\end{align}
A: Here is a proof of "only if":
$$|a|=\frac{1}{2}|2a|=\frac{1}{2}|(z+a)-(z-a)|
  \le\frac{1}{2}\bigl(|z+a|+|z-a|\bigr)=|c|\ .$$
The converse is trivial if $a=0$, otherwise let $b$ be the real number
$$b=\sqrt{\frac{|c|^2}{|a|^2}-1}$$
and consider $z=abi$.
