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

• You only showed that $$|z-1|+|z+1|\leqslant 2\implies x^2+y^2\leqslant1.\tag{z=x+iy}$$ – Workaholic Feb 3 '16 at 15:21

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.

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.

• I am not sure I follow what you said. I only saw a triangle inequality couple of minutes ago, as it appeared in the book I am doing exercises from (Complex Analysis by Theodore W. Gamelin). Would it be too much to ask how do you prove it the other direction? – Scavenger23 Feb 3 '16 at 15:33
• @KuderaSebastian Sure. 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 – Stella Biderman Feb 3 '16 at 15:36