Solve for $z$ in equation $|z-a|+|z+a|=2|c|$

Question

I'm trying to solve for $z$ in the equation $$|z-a|+|z+a|=2|c|.$$ My idea is to square both sides

$$|z-a|^2+|z+a|^2+2|(z-a)(z+a)|=4|c|^2$$

Using $|z|^2=z\bar{z}$, this becomes

$$|z|^2+|a|^2+|(z-a)(z+a)|=2|c|^2$$

I'm not sure how to continue from here.

What do you mean by "solve for $z$"? The set of solutions is an ellipse, after all :) — Ted Shifrin, Dec 15 '13 at 16:31
@TedShifrin Oh, I thought maybe there's some way to simplify this further. But yes, now I realize it's an ellipse, so it's already in the best form. :) — Mika H., Dec 15 '13 at 16:34

score 1 · Accepted Answer · answered Dec 16 '13 at 05:19

You can write down all solutions in parametric form. If $|a|=|c|$, they are $z=ta$, $-1\le t\le 1$. If $|a|<|c|$, we have an ellipse with focal distance $2|a|$ and major axis $2|c|$. The minor semiaxis is $\sqrt{|c|^2-|a|^2}$. Hence, the parametric equation is $$ z = \frac{|c|}{|a|} a \cos t + \frac{\sqrt{|c|^2-|a|^2}}{|a|} ia\sin t,\quad 0\le t\le 2\pi $$ This works when $|c|=|a|$, too.

Solve for $z$ in equation $|z-a|+|z+a|=2|c|$

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