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

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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.

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    @Adam: To say that a set $S$ is unbounded just means that for each real c>0 there is a $z\in S$ such that $|z|\ge c$. To show that $S$ is unbounded, you simply show that for each c>0 there is at least one such $z$. If $S$ is the set consisting of the two coordinate axes, the real numbers $c$ and $-c$ and the imaginary numbers $ci$ and $-ci$ all satisfy that inequality and are in $S$. (Of course any real or pure imaginary with a magnitude greater than $c$ also works>02012-11-04