# Proving that a complex set in open/closed/neither and bounded/not bounded

The set $$\{z\in\Bbb C:1<|2z-6|\le 2\}$$
and the set $$\{z\in\Bbb C:|z|=|\Re(z)|+|\Im(z)|\}$$

#### Solutions Collecting From Web of "Proving that a complex set in open/closed/neither and bounded/not bounded"

Your first step should be work out what the sets look like. I’ll leave the first one to you and get you a good start on the second one.

If $z=x+iy$, then $|z|=\sqrt{x^2+y^2}$, $|\Re(z)|=|x|$, and $|\Im(z)|=|y|$, so

$$\{z\in\Bbb C:|z|=|\Re(z)|+|\Im(z)|\}=\left\{x+iy\in\Bbb C:\sqrt{x^2+y^2}=|x|+|y|\right\}\;.$$

The defining equation on the righthand side can be squared to give $x^2+y^2=\left(|x|+|y|\right)^2$, or, after multiplying out, $x^2+y^2=x^2+2|xy|+y^2$; it should be an easy matter to use this to see what the set actually looks like, and once you’ve done that, it should be clear that the set is closed and unbounded.