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Question on graphing in the complex plane

  1. Jul 5, 2009 #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?

  2. jcsd
  3. Jul 5, 2009 #2
    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.
  4. Jul 6, 2009 #3


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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.
  5. Jul 6, 2009 #4
    Does this look right? Including the the 5 sets I had to determine.


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

    Attached Files:

    Last edited: Jul 6, 2009
  6. Jul 7, 2009 #5
    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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