Describe the set of all complex numbers $z$ such that $|z-a |+| z-b |=c$

Question

Describe the set of all complex numbers $z$ such that : $$|z-a |+| z-b |=c$$ where $a,b, c$ are real

At a simple look I immediately recognized that this is some ellipse because it's the same with the definition of the ellipse with two foci, the sum of the distances from a point in the ellipse in this case $z$ remains constant. I started working out on algebraic manipulations based on the rectangular coordinates of $z$ but the equation gives me only square roots like:

$$\sqrt{(x-a)^{2}+y^2}+\sqrt{(x-b)^2+y^2}=c$$

but this is at least not very convincing, can someone give some hint how to prove this more analytically.

Is $a$ a real or generic complex number? Why are you ignoring the imaginary part of $a$? — Marco Disce, Feb 27 '16 at 09:24
A very analytical way would be to say that |z-x| represents the distance of z from x, so $$|z-a|+|z-b|=c$$ represents the locus of a point which moves such that the sum of the distance from two points (a and b) remains constant (i.e c) which represents an ellipse — Nikunj, Feb 27 '16 at 09:31

Jean Marie · Accepted Answer · 2016-02-27T10:10:28.200

2

If $|b-a|<c$, the locus is an ellipse with foci in $a$ and $b$.

If $|b-a|=c$, it is the line segment $[a,b]$.

If $|b-a|>c$, it is the void set.

Edit 1 (Commentaries) :

For me, the problem has to be converted into: find the locus of points such that d(M,A)+d(M,B)=constant. If you want to transform it into analytical charactrization in $\mathbb{R}^2$, you will have to square both sides of your equation ; it will remain a central term with a square root, you will have to isolate it, say in the right hand side of the equation, square both sides again ; in this way you end up with a fourth degree equation that you have to simplify in order to lower it to the second degree equation of an ellipse. Wow! Do you think this is what is asked to you ?

Edit 2: I had interchanged the "<" and the ">". It is corrected now.

edited Feb 27 '16 at 10:10

answered Feb 27 '16 at 09:42

Jean Marie

81,803

isn't this quite short? and why have you divided it in cases? – Admir Kullolli Feb 27 '16 at 09:52
See the Commentaries I have place in my first answer. – Jean Marie Feb 27 '16 at 10:05
Sorry, I had interchanged the "<" and the ">". It is corrected now. – Jean Marie Feb 27 '16 at 10:12
can you give me some references where to look up these conditions for the ellipse? – Admir Kullolli Feb 27 '16 at 18:29
why it is void set when $|b-a| \gt c$, suppose if we take $|z-1|+|z+1|=0.5$, then we get the Locus as $15x^2-y^2+8x+\frac{9}{16}=0$ clearly it is a hyperbola. Geometrically i do not know what exactly happens here – Umesh shankar Jul 06 '17 at 01:23

score 1 · Answer 2 · answered Feb 27 '16 at 10:17

1

Hint:

$$|z-a|+|z-b|=c$$

Square

$$|z-a|^2+2|z-a||z-b|+|z-b|^2=c^2.$$

Move the squares from the LHS to the RHS and square again

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

Then,

$$0=c^4+|z-a|^4+|z-b|^4\color{red}{-2|z-a|^2|z-b|^2}-2c^2|z-a|^2-2c^2|z-b|^2\\ =c^4+(|z-a|^2-|z-b|^2)^2-2c^2|z-a|^2-2c^2|z-b|^2.$$

Fully developing the last expression, you get the equation of a quadric ($x^2$ and $y^2$ cancel out in the parenthesis). The terms of the second degree are

$$(2(x_a-x_b)x+2(y_a-y_b)y)^2-4c^2(x^2+y^2)\\ =4((x_a-x_b)^2-c^2)x^2+8(x_a-x_b)(y_a-y_b)xy+4((y_a-y_b)^2-c^2).$$

The type of the conic depends on the sign of

$$\Delta=(x_a-x_b)^2(y_a-y_b)^2-((x_a-x_b)^2-c^2)((y_a-y_b)^2-c^2)\\ =c^2((x_a-x_b)^2+(y_a-y_b)^2-c^2)=c^2(|a-b|^2-c^2).$$

answered Feb 27 '16 at 10:17

One more thing , in my answer I have : If $|b-a|<c$, the locus is an ellipse with foci in $a$ and $b$.
If $|b-a|=c$, it is the line segment $[a,b]$.

If $|b-a|>c$, it is the void set.
– Admir Kullolli Feb 27 '16 at 18:16
Why do we have these conditions? – Admir Kullolli Feb 27 '16 at 18:17
Have alook at http://www.thisiscarpentry.com/2013/09/06/drawing-an-ellipse-the-string-method/ The length of the rope has to be larger than the distance between the pins... – Jean Marie Feb 27 '16 at 18:22
@AdmirKullolli: this obvious from my answer. You were asking for an analytical solution and accepted a trivial one. Grrr. – Feb 29 '16 at 07:12
I think there is a mistake in your calculations that's why I didn't upvote your answer, maybe I am wrong, thanks a lot anyways – Admir Kullolli Feb 29 '16 at 08:30
@AdmirKullolli: please point the mistake. – Feb 29 '16 at 09:25
At the red part, from my calculations it should be -6 not -2 – Admir Kullolli Feb 29 '16 at 15:55
@AdmirKullolli: the red part is correct, $2-4=-2$, review your calculation. – Feb 29 '16 at 15:58
Ok you are right, my mistake, can we chat for another equality I am interested, I have a solution but I would like to consult it with you – Admir Kullolli Feb 29 '16 at 16:08
@AdmirKullolli: sorry, no, I don't work on demand. Just post it. – Feb 29 '16 at 16:54
Ok, sorry @YvesDaoust – Admir Kullolli Feb 29 '16 at 16:55

Describe the set of all complex numbers $z$ such that $|z-a |+| z-b |=c$

2 Answers2

Linked