Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.