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Fun with sets

  1. Mar 20, 2006 #1
    For the following open sets
    a) |Arg z| <pi/4
    b) -1 < Im z <= 1
    c) (Re z)^2 > 1

    Sketch the sets
    Which are domains
    Describe the boundary

    a) |Arg z |< pi/4

    well ok that means
    [tex] |\theta_{0} + 2 \pi k| < \pi/4 [/tex]
    so the theta must alwasy be less than pi/4
    im not sure how to sketch this... i mthinking its the attached diagram, but only the triangle in the first quadrant
    yes it is a domain
    umm is there a way to do with without diagram??

    b) -1 < Im z <= 1
    shaded part of 2
    yes it is a domain
    the region bounded by y = -1 from the bottom and unbounded above

    c) (Re z)^2 > 1
    like 3?
    not a domain

    Attached Files:

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  2. jcsd
  3. Mar 20, 2006 #2
    Is this the principal argument: Arg as opposed to arg? If so then where is the branch cut for Arg? And what does that absolute value sign do?
  4. Mar 20, 2006 #3
    it is Arg
    what do u mean branch cut??
  5. Mar 20, 2006 #4
    How is Arg z (the principal argument of z ) defined?
  6. Mar 20, 2006 #5
    the branct cut is at pi/2 for the principle argument yes??
  7. Mar 21, 2006 #6
    No... And if it were the question posed to you would be somewhat impossible.
  8. Mar 21, 2006 #7


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    Science Advisor

    [tex]|Arg z|< \frac{\pi}{4}[/tex]

    It really doesn't matter whether or not "Arg" means the principal argument: since it's between [itex]-\frac{/pi}{4}[/itex] and [itex]\frac{\pi}{4}[/itex], it IS, except for the negative values, the principal argument!
    Stunner5000pt, it's not just in the first quadrant- since this is absolute value, you can have negative values of the argument and so include the fourth quadrant.

    "b) -1 < Im z <= 1
    shaded part of 2
    yes it is a domain
    the region bounded by y = -1 from the bottom and unbounded above"

    Why unbounded above? It say "Im z<= 1".

    "c) (Re z)^2 > 1"

    In other words, Re z> 1 or Re z< -1.
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