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