Question on graphing in the complex plane

  • #1
Okay, I need to graph the following set in the complex plane:

M={z[tex]\in[/tex]C:[(1<|z-i|[tex]\leq[/tex]2) and (z[tex]\neq[/tex]2+i)] or [z = 1 + [tex]\pi[/tex]i]}


I got the last two constraints, but the first one is what's giving me trouble.

is z-i just x+yi that is (1,1) on the complex plane lowered by 1?


Thanks
 

Answers and Replies

  • #2
537
3
The region described by [itex]1<|z-i|\leq 2[/itex] is the annulus centered at z=i with the inner radius 1 and outer radius 2 and you include the circle |z-i|=2 since you have a less than or equal sign.
 
  • #3
HallsofIvy
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Just to expand on n!kofeyn's response: |z| is the distance from 0 to z and so |z- a| is the distance from a to z. |z- i| is the distance from z to i. If |z-i|= 1, the distance from i to z is always 1 so this is a circle of radius 1 centered on i. If |z-i|= 2, this is circle of radius 2 centered on i. [itex]1< |z-i|\le 2[/itex], then, is the points between the two circles, an "annulus". Because |z-i|= 1 is NOT included, the inner boundary of the annulus is not included in the set but because |z- i|= 2 is included, the outer boundary of the annulus is part of the set.
 
  • #4
Does this look right? Including the the 5 sets I had to determine.

Thanks!

PS I know it says Final Exam at the top, but I'm not cheating or anything its one from last year :)
 

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  • #5
537
3
Your picture is a little fuzzy, but it looks correct. You need a less than or equal to sign to include 1 for the set of limit points. The boundary |z-i|=1 is part of the set of limit points. Also, your complement is a little wrong. You are union-ing all the points that are not 1+pi*i, which is everything in the complex plane minus that one point! So fix your limit points and complement, and this solution should be good.
 

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