# Set sketch of $z$ element of $\mathbb{C}$ such that $1 < |2z - 6| \le 2$

How would I do this?

I know it's a circle of radius half around 3 and a larger circle of radius 1 around 3. But why?

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It's not. It's a half-open annulus. en.wikipedia.org/wiki/Annulus_%28mathematics%29 – Fly by Night Nov 3 '12 at 15:19
Is this a homework problem? I suspect so, since another user asked another question about precisely this set (and another) today. If so, please add the homework tag. Don't worry, people will still help you! – Cameron Buie Nov 3 '12 at 18:40

With complex numbers, $|a-b|$ will give you the distance between $a$ and $b$. Hence $$|2z-6|=2|z-3|$$ is twice the distance from $z$ to $3$. Divide your inequalities by $2$, and you get $${1\over 2}<|z-3|\leq 1$$ In other words, the distance between $z$ and $3$ should lie between $1\over 2$ and $1$, which is the same as the answer you gave.
 Ah, yes, that's what I was thinking, it's the larger circle which confuses me radius 1 around 3? – Rebecca Shaw Nov 3 '12 at 15:15 Is the larger circle before you divide by 2? I need to have both so I can say whether it's bounded, not or either and whether it's open, closer or neither. – Rebecca Shaw Nov 3 '12 at 15:17 I'm not sure I understand your question. Your inequalities $1<|2z-6|\leq 2$ is equivalent to my inequalities ${1\over 2}<|z-3|\leq 1$, i.e., they have the same solution. – Per Manne Nov 3 '12 at 15:19 Oh nevermind, I've just realised I've been an idiot ha. Thanks for your help – Rebecca Shaw Nov 3 '12 at 15:20 It's worth noting that the $z$ in the set may lie on the larger circle, but cannot lie on the smaller one. – Cameron Buie Nov 3 '12 at 18:42