# Geometrically interpret the set $\{z \in \Bbb C :$ $|z+a|+|z-a| \leq 2b,$ $b\in\Bbb R^+,$ $|a|<b\}.$

Geometrically interpret and determine the following set of complex numbers: $$\{z \in \Bbb C : |z+a|+|z-a| \leq 2b , b\in\Bbb R^+,|a|<b\}.$$

I understand it means the sum of distance between $z$ and $a,$ and between $z$ and $-a$ is less than or equal to $2b.$ I drew a circle with centre at origin and with radius $b,$ a lies within the circle, $-a$ is the reflection of a across the origin, and I'm stuck.

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Presumably, you want the conditions on $a$ and $b$ outside the set definition:$$\{z\in C: |z+a|+|z-a|\leq 2b\}$$ where $b\in\mathbb R^+$ and $|a|<b$. – Thomas Andrews Aug 17 '12 at 18:52
Is $a$ strictly real or complex? $|a|<b$ is ambiguous. – Taylor Martin Aug 17 '12 at 20:43

Hint: you need to know one fact from traditional geometry, which is the fact that if a point moves so that the sum of its distances from two fixed points is constant, then the locus is an ellipse, and the two fixed points are the foci of the ellipse.

Wikipedia has some details here.

For your example, the "less than or equal to" just gives the interior of the ellipse.

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@Old John's traditional geometry answer is obviously all right and, well... geometric, but, if one needs to be reassured, the Cartesian brute force approach works as well.

To see this, assume without loss of generality that $a$ is a real number, then $z=x+\mathrm iy$ yields $|z\pm a|=\sqrt{x^2\pm2ax+a^2+y^2}$. Squaring this, one sees that the desired set has equation $$x^2\pm2ax+a^2+y^2=|z\pm a|^2\leqslant(2b-|z\mp a|)^2=4b^2+x^2\mp2ax+a^2+y^2-4b|z\mp a|,$$ that is, $$b^2\mp ax\geqslant b|z\mp a|,$$ or, squaring again, $$b^4\mp2b^2ax+a^2x^2\geqslant b^2(x^2\mp2ax+a^2+y^2),$$ that is, $$(b^2-a^2)x^2+b^2y^2\leqslant b^2(b^2-a^2),$$ which describes the interior of an ellipse with the coordinate axes as principal axes and the origin as center. Equivalently, $$\frac{x^2}{b^2}+\frac{y^2}{c^2}\leqslant1,\qquad\text{with}\quad c^2=b^2-a^2.$$

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Ah yes, but the title does explicitly state "geometrically" ... :) – Old John Aug 17 '12 at 19:23
@OldJohn Indeed... :-) – Did Aug 17 '12 at 19:25

A really quick way to "interpret" (in the sense of pictures, not analytic precision) the locus of points is to realize that taken by themselves, the inequalities $$|z-a|\leq2b$$ and $$|z-(-a)|\leq2b$$ represent discs with centers $a$ and $-a$, respectively. Because of this symmetry, and the condition that $|a|<b$, the discs have a non-empty intersection which intersects in an ellipse whose major and minor axes intersect at an equilateral point with respect to the position of $a$ in the complex plane (this is just a fancy way to say the ellipse has a center at the origin since I assume $a$ is real). (To get the expression of the ellipse is not difficult; see above answers). If you are having trouble imagining this, just think of a prototypical ven diagram.

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the discs have a non-empty intersection... OK, which intersects in an ellipse whose major and minor axes... Sorry? What is an intersection which intersects in something? – Did Aug 24 '12 at 15:40

put z=x+iy |x+iy+a|+|x+iy-a|=2b root((x+a)^2+y^2 )+root((x-2)^2+y^2 )=2b it is ellipse

root((x+a)^2+y^2 )+root((x-2)^2+y^2 )<2b area inside the ellipse

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