Absolute value equality involving complex analysis I'm preparing for a complex analysis prelim which I'll take next summer by consulting Ahlfor's Complex Analysis: An Intro to the Theory of Analytic Functions of One Variable (3rd edition). My question pertains to Exercise 4 of  Chapter 1, Section 1.5, and it is as follows:  
Show that there are complex numbers $z$ satisfying $$|z-a| + |z+a|= 2|c|$$ if and only if $|a| \leq |c|$. If this condition is fulfilled, what are the smallest and largest values of |z|?  
Well, for the first part of the question I was able to show only one direction, i.e. if there are complex numbers $z$ such that $|z-a| + |z+a|= 2|c|$, then $|a| \leq |c|$ by applying the triangle inequality. Specifically, I did the following: $$ \begin{array} {lcl} 2|a| &=& |2a| \\ &=& |a-z+z+a| \\&\leq& |a-z| + |z+a| \\&=& |z-a| + |z+a| \\&=& 2|c|. \end{array}$$ Dividing both sides of the inequality by $2$ yields the result in one direction. The other direction is giving me trouble, and so is the second part of the exercise. Some helpful hints and advice would be much appreciated.

You can rotate the complex plane so that $a$ is real.
Let $z= x+ i \,y$. Then the left hand side is 
$$
\sqrt{(x+a)^2+y^2} + \sqrt{(x-a)^2+y^2}$$
is minimum when $y=0$, i.e when $z$ is real. Clearly all $z$ in the interval $[-a,a]$ satisfies the condition.
Hence $z=0$ is the minimum, $z=\pm a$ is maximum in magnitude
Note: Added as an after thought.
If you do not want to make the initial rotation, then just consider $z$ of the form
$$
z = \lambda \, (-a) + (1-\lambda) a = a \,(1 - 2 \lambda),~~~ 0 \le \lambda \le 1$$
and show that all these $z$ meet the requirement.
