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## Homework Statement

Identify the set of points satisfying ##1<\vert 2z-6\vert <2## such that ##z\in\Bbb{C}##.

My pre-caculus is very rusty, so I am not sure if I am doing this correctly.

## Homework Equations

##x^2 +y^2= r^2##

##\forall z,z'\in\Bbb{C}, \vert zz'\vert =\vert z\vert\vert z'\vert##

## The Attempt at a Solution

Let ##z=x+iy##. Then ##\vert 2z-6\vert =\vert 2(z-3)\vert=\vert 2\vert\vert z-3\vert## and \begin{align}1<\vert 2z-6\vert<2\Longleftrightarrow \frac{1}{2}<\vert z-3\vert <1\end{align}

We want complex numbers ##z## that are at least ##1/2## units away from ##3## and at most ##1## unit away from ##3##. Let ##H=\{z\in\Bbb{C}:\frac{1}{2}<\vert z-3\vert\}## and ##O=\{z\in\Bbb{C}:\vert z-3\vert <1\}##. We want to find the conditions of all ##z\in\Bbb{C}## such that ##z\in H\cap O##. Since ##\Bbb{R}^2\cong\Bbb{C}##, then the general form of elements in ##H## can be found by solving: ##(x-3)^2 +y^2>(\frac{1}{2})^2\Longrightarrow y>\pm\sqrt{\frac{1}{4}-(x-3)^2}## and for ##O## we solve: ##(x-3)^2+y^2<1^2\Longrightarrow y<\pm\sqrt{1-(x-3)^2}##. Therefore, we have \begin{align}\pm\sqrt{\frac{1}{4}-(x-3)^2}<y< \pm\sqrt{1-(x-3)^2}\Longleftrightarrow 1/4<y<1\end{align}

Hence, ##H\cap O=\{z=x+iy\in\Bbb{C}:\pm\sqrt{\frac{1}{4}-(x-3)^2}<y< \pm\sqrt{1-(x-3)^2}\quad \land\quad 1/4<y<1 \}##