For the equation $$|z-a|+|z+a|=2|c|,$$ where $|a|\le|c|$, what are the smallest and largest values of $|z|$?
I am finding various bounds, but I don't know which bounds are the most precise ones. For example, I found that $|z| \le |c|$, but can this bound be improved?
Using arithmetic-quadratic mean inequality, $$|c| = \frac{|z-a|+|z+a|}{2} \le \sqrt{\frac{|z-a|^2+|z+a|^2}{2}}=\sqrt{|z|^2+|a|^2}.$$ Consequently, $|z| \ge \sqrt{|c|^2-|a|^2}$. The minimum value is achieved when $z = \pm i a \sqrt{\frac{|c|^2}{|a|^2}-1}$.
For the maximum value of $|z|$, observe that $$2|c|=|z-a|+|z+a| \ge |z+a+z-a|=2|z|.$$ Consequently, $|z|\le|c|$, and the maximum value is achieved when $z = \pm a \frac{|c|}{|a|}$.
Geometric hint: the locus $|z-a|+|z+a|=2|c|$ of points $z$ with a constant sum of distances to two fixed points $a,-a$ is an ellipse centered at the origin, with foci $-a, a$ and semi-major axis $|c|\,$.
The point farthest from the origin is either vertex $z=\lambda \cdot a\,$, which substituting into the equation gives $\lambda=|c|/|a|\,$, so the vertex is $z = |c|/|a| \cdot a\,$ and the maximum magnitude is $|z| = |c|\,$.
The point closest to the origin is either co-vertex $z=i\,\mu \cdot a\,$, which substituting into the equation gives $\mu=\sqrt{|c|^2/|a|^2-1}\,$, so the co-vertex is $z = i\,\sqrt{|c|^2/\,|a|^2-1} \cdot a\,$ and the minimum magnitude is $|z| = \sqrt{|c|^2 - |a|^2}\,$.
