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 October 1st, 2016, 07:57 AM #1 Newbie   Joined: Oct 2016 From: Lawrence, Kansas Posts: 2 Thanks: 0 graphing set How do I graph a set that is defined as z given that |z -[complex number]| <= 1 (apologies: I am approaching an online class with an undeniably weak background, but with determination....). Last edited by skipjack; October 22nd, 2016 at 10:31 AM.
 October 1st, 2016, 08:06 AM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,481 Thanks: 744 You'll see the expression $|z - z_0|$ a lot. The region $|z-z_0| \leq \rho$ is a disk on the complex plane of radius $\rho$, centered at the point $z_0$. Thanks from Dan dimmitt Last edited by skipjack; October 22nd, 2016 at 10:32 AM.
 October 1st, 2016, 08:27 AM #3 Newbie   Joined: Oct 2016 From: Lawrence, Kansas Posts: 2 Thanks: 0 Thanks! Last edited by skipjack; October 22nd, 2016 at 10:31 AM.
 October 22nd, 2016, 09:38 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,714 Thanks: 697 Specifically, in the "complex" plane z= x+ iy. Writing point $\displaystyle z_0$ as a+ iy, then $\displaystyle |z- z_0|= |x+ iy- (a+ ib)|= |(x- a)+ i(y- b)|= \sqrt{(x-a)^2+ (y- b)^2}$ so that $\displaystyle |z- z_0|\le \rho$ is $\displaystyle \sqrt{(x- a)^2+ (y- b)^2}\le \rho$ and squaring both sides, $\displaystyle (x- a)^2+ (y- b)^2\le \rho^2$. $\displaystyle (x- a)^2+ (y- b)^2= \rho^2$ is the circle, in the complex plane with center $\displaystyle z_0= a+ ib$ and radius $\displaystyle \rho$.

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