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Let $a$ and $c$ be complex numbers. Show there exists complex numbers $z$ s.t. $|z-a|+|z+a| = 2|c|$ if and only if $|a| \leq |c|$.

I've shown the forward direction correctly, but I don't know how to get the backwards direction.

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2 Answers 2

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Let $z = \frac{a|c|}{|a|}$ for $|a| \neq 0$ and $|c|\geq|a|$ (The case $|a|=0$ is trivial). Then, \begin{align} |z-a|+|z+a| & = \left|\frac{a|c|}{|a|}-a\right| + \left|\frac{a|c|}{|a|}+a\right| \\ & = |a|\left(\left|\frac{|c|}{|a|}-1\right| + \left|\frac{|c|}{|a|}+1\right|\right) \\ & = |a|\left(\frac{|c|}{|a|}-1 + \frac{|c|}{|a|}+1\right) \\ & = 2|c|. \end{align}

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  • $\begingroup$ So for the trivial case, you can't use the quotient you've set $z$ equal to? $\endgroup$ Commented Apr 25, 2013 at 1:46
  • $\begingroup$ @AlanH $z=c$ works if $|a|=0$ (or equivalently if $a=0$). $\endgroup$
    – Lord Soth
    Commented Apr 25, 2013 at 1:58
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$|z-a|+|z+a| = 2|c|$ represents an ellipse, if $|c| < |a|$ then what would happen?

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  • $\begingroup$ I'm aware of that answer, but I'm looking for an algebraic answer simply because that wouldn't come to me immediately. How did you see that it was an ellipse? I ask because it's not in the familiar form that you see in conic sections of calculus. $\endgroup$ Commented Apr 23, 2013 at 22:20
  • $\begingroup$ @Alan: An ellipse is the set of points in the plane such that the sum of the distances to two specific points--called the ellipse's foci--is some given number. (A circle is a special case of an ellipse, where the foci coincide.) In this particular case, it is the ellipse with foci $\pm a$, such that the sum of the distances to the foci is $2|c|$. $\endgroup$ Commented Apr 24, 2013 at 4:15

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