Prove that $|Re(z_{1})-Re(z_{2})|\leq |z_{1}-z_{2}|\leq |Re(z_{1})-Re(z_{2})|+|Im(z_{1})-Im(z_{2})|$.

Question

Prove that $|Re(z_{1})-Re(z_{2})|\leq |z_{1}-z_{2}|\leq |Re(z_{1})-Re(z_{2})|+|Im(z_{1})-Im(z_{2})|$.

I understand how to get $|z_{1}-z_{2}|\leq |Re(z_{1})-Re(z_{2})|+|Im(z_{1})-Im(z_{2})|$

and $|Re(z_{1})-Re(z_{2})|\leq |Re(z_{1})-Re(z_{2})|+|Im(z_{1})-Im(z_{2})|$.

But what about $|Re(z_{1})-Re(z_{2})|\leq |z_{1}-z_{2}|$? Do I use the reverse triangle inequality?

Also: Does this mean that $|Im(z_{1})-Im(z_{2})|\leq |z_{1}-z_{2}|$?

Answer 1

The statement is equivalent to $$|{\rm Re}(z)|\leq |z|\leq |{\rm Re }(z)|+|{\rm Im}(z)|,$$ where $z=x+yi,x,y\in {\mathbb R}.$ This is true because$$|x|\leq \sqrt{x^2+y^2}\leq |x|+|y|.$$

Answer 2

For complex $w$, $$|w|^2 = (\Re(w))^2 + (\Im(w))^2 \ge |\Re(w)|^2.$$ Your inequality is this but with $w = z_1 - z_2$.