<s>[INST] Describe the set of all complex numbers $z$ such that $|z-a |+| z-b |=c$ 
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.
 [/INST] 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
$$\Delta=(x_a-x_b)^2(y_a-y_b)^2-((x_a-x_b)^2-c^2)((y_a-y_b)^2-c^2)\\
=c^2((x_a-x_b)^2+(y_a-y_b)^2-c^2)=c^2(|a-b|^2-c^2).$$
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