Homework: Restriction on |z+1|+|z-1|=2, z∈C

Question

After simplifying the relation $|z+1| + |z-1| = 2, z∈\mathbb{C}$ to $\Im(z)=0$, I plotted the original relation on my TI-nspire calculator and WolframAlpha. As expected it simplifies to $\Im(z)=0$ however both stated that the relation exists where $-1 < Re(z) < 1$. Can someone explain how this restriction can be found when simplifying the relation?

Here's a link to the relation plotted on WolframAlpha.

Answer 1

By the triangle inequality: $2=|z+1|+|z-1|\geq |z+1+z-1|=2|z|$.

Hence we know even before doing any more sophisticated computation that $|z|\leq 1$, which when $z$ is real translates into the restriction you mention.

Answer 2

An alternative way you have that $$-1\leq\Re(z)+\Im (z)\leq1$$ Since $$|z+1|+|z-1|=\sqrt{(\Re(z)+1)^2+\Im(z)^2}+\sqrt{(\Re(z)-1)^2+\Im(z)^2}$$ Now if $$\Re(z)>1$$ than $$2\leq\sqrt{(1+1)^2+\Im(z)^2}<\sqrt{(\Re(z)+1)^2+\Im(z)^2}$$ and if $$\Re(z)<-1$$ than $$2\leq\sqrt{(-1-1)^2+\Im(z)^2}<\sqrt{(\Re(z)-1)^2+\Im(z)^2}$$