How do I identify the set of points satisfying $|z-1|+|z+1|\leq 2$? How do I identify the set of points satisfying $|z-1|+|z+1|\leq 2$?
My idea is:     $|z+1|^{2}=|z|^2+1+2x$
$|z-1|^{2}=|z|^2+1-2x$
$|z-1|+|z+1|\leq 2$ $\implies$ $|z+1|\leq 2 -|z-1|$
$(|z+1|)^2\leq (2 -|z-1|)^2$ (* is this step correct?)(because $-3<2$ but $9 \nleq 4$)
I am confused at this step of my solution. Can anyone suggest me how do I fix this problem?
 A: Think of the triangle inequality.
Since the distance from $z=-1$ to $z=1$ is $2$,
the only points satisfying the (in)equality are on the line segment from $z=-1$ to $z=1$.
A: I agree with J. W. Tanner's solution, and would like to offer two alternative approaches, both less elegant than his.
Given any point $(x,y) = (x + iy),\;$
let $S[(x,y)]$ denote the sum of its distances from the two points 
(-1, 0) and (1,0). 
That is, $S[(x,y)] = |(x + iy) - 1| + |(x + iy) + 1|.$ 
It is desired to find all $(x,y) \;\ni \;S[(x,y)] \leq 2.$
$\underline{\text{Solution 1:}}$
In the complex plane, construct two circles, each of radius 1, with one circle centered at (1,0) and the other circle centered at (-1, 0).  It is immediately obvious that if $(x,y)$ is outside both circles, then its distance to each center will be $> 1,$ which will imply that $S[(x,y)] > 2.$
Therefore, the search for satisfying points $(x,y)$ can be confined to those points that are on or interior to at least one of the two circles.  Assume therefore that $(x,y)$ is on or interior to one of the two circles.  Viewing the distance of $(x,y)$ to each of the centers (1,0) and $(-1,0)$, the following geometric conclusions are immediate:
(1) $y \neq 0 \;\Rightarrow\; S[(x,y)] > S[(x,0)].$ 
(2) $\;$ For $-1 \leq x \leq 1, \;S[(x,0)] = 2.$ 
(3) $\;$ For either $x < -1$ or $x > 1, \;S[(x,0)] > 2.$
$\underline{\text{Solution 2:}}$
$S[(x,y) \leq 2 \;\Rightarrow\; 
|(x - 1) + iy| \leq 2 - |(x + 1) + iy| \;\Rightarrow\;$ 
$(x-1)^2 + y^2 \;\leq\; 4 + (x+1)^2 + y^2 \;-\;
4 \sqrt{(x + 1)^2 + y^2} \;\Rightarrow\;$ 
$4 + 4x \geq 4 \sqrt{(x + 1)^2 + y^2} \;\Rightarrow\;$ 
$(x + 1) \geq \sqrt{(x + 1)^2 + y^2} \;\Rightarrow\;$ 
$(x + 1)^2 \geq (x + 1)^2 + y^2.$
At this point, it is immediate that only equality is achievable, and that only when $y = 0.$
Addendum
I can't help wondering what the intended solution is.  My initial strong instinct is
that J.W.Tanner's answer was intended, because the math student is supposed to
"build on recent results", and each of my two solutions can be interpreted as
"re-inventing the wheel" of the core concept of J.W. Tanner's answer.
On the other hand, both of my solutions were facilitated because (-1,0) and (1,0)
have the same imaginary component.  Suppose that the problem had instead been something like:
find all satisfying $z \;\ni \;|\,z - [(-1) + i(0)] \,| \;+\; |\,z - [(2) + i(4)] \,| \;\leq 5.$
With this alternate problem, J.W.Tanner's approach is tailor made, once it is recognized
that 
$[(-1,0)] ~:~ [(2,0)] ~:~ [(2,4)]$ form a 3-4-5 right triangle.
With this alternate problem, either of my two solutions would have been significantly more
difficult to use.
A: $|z+1|+|z-1|=2$ denotes a line segment. But $|z+1|+|z-1|<2$ is a null set and it represents nothing in the Argand plane.
And when $|z+1|+|z-1|>2$, represents region (point) out side  an ellipse.
