Sketch the set of points satysfing an inequality $|z+1|+|z-1|\leq2$ The inequality is 
$$|z-1|+|z+1|\leq2$$
I used a triangle inequality to show that 
Since triangle inequality states:
$$|z+w|\leq|z|+|w|$$
Then
$$|z-1+z+1|\leq|z-1|+|z+1|\leq2$$
So $$|2z|\leq2$$
From this point I expanded out
$$\sqrt{(2x)^2+(2y)^2}\leq2$$
Or $$4x^2+4y^2\leq4$$
So I end up with a disk centred at origin, radius 1. However my "answer" at the back of the page says "one point" (it doesnt give step by step solution)Thanks for the help
 A: Your book is clearly wrong, as both $-1$ and $1$ are solutions to this equation. You are also wrong however, as plugging in $i$ gives us $2\sqrt{2}$.
It should be clear that $a\in [-1,1]$ is a solution and no other real number is. Now consider $a+bi$. The value of the LHS is equal to $2$ for every $a\in[-1,1]$. Thus by the triangle inequality, if $b\neq 0$ we would get a value greater than $2$ since both absolute values would increase in size.
This is because when you plug in $a$, a real number, to the inequality you have, you get $2=2$. If we consider $a+bi$, then $$|(a-1)+bi|\geq |a-1|+|bi|>|a-1|$$ and likewise for the second absolute value. Thus for any nonreal value, the LHS is greater than 2
The mistake you made was the direction you proved. You proved that if $z$ satisfies the inequality, then it falls in the unit disc. This is true, but gives rise to false positives. You need to prove that if $z$ falls on the unit interval, then it satisfies the inequality.
A: Solutions on the real line lie in the closed interval $[-1,1]$. Points outside this interval give a result greater than $2$. Points in the interval give the value $2$.
Off the real line you can use the triangle inequality to show that the distances between $z$ and the two points $+1$ and $-1$ add to strictly more than the the value for the point on the real line which has value the real part of $z$ (carefully chosen segments on the real line make the base of the triangle). Draw a diagram.
