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Complex analysis: Sketch the region in the complex plane

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Sketch:
    {z: [itex]\pi[/itex]?4 < Arg z ≤ [itex]\pi[/itex]}

    2. Relevant equations



    3. The attempt at a solution
    Is it right to assume

    z0 = 0 ; a = a (radius = a) ; and taking [itex]\alpha[/itex] = [itex]\pi[/itex]/4 ; [itex]\beta[/itex] = [itex]\pi[/itex]

    And now in order to sketch the problem after setting up the complex plane is it correct to to plot z0 at the origin and then from the origin plot [itex]\pi[/itex]/4 by rotating to the right in a clockwise rotation for [itex]\pi[/itex]/4 radians for the first condition and then rotating [itex]\pi[/itex] to the left from the origin (anti-clockwise rotation) for the second condition and then using a solid or dashed line according to the strictly < or ≤ conditions and this gives me the correct region?

    Basically I am confused as to how to rotate the angle in terms of clockwise or anti-clockwise according to the conditions given.
     
  2. jcsd
  3. Mar 4, 2012 #2
    And I am also unsure if my radius is in fact a or am I missing an important step?
     
  4. Mar 4, 2012 #3

    tiny-tim

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    Hi Rubik! :smile:

    do you mean {z: [itex]\pi[/itex]/4 < Arg z ≤ [itex]\pi[/itex]} ?
    no, everything is always anti-clockwise

    i'm worried why you thought it wasn't :confused:

    (and i don't understand where radius comes into it)
     
  5. Mar 4, 2012 #4
    Oops yep I meant [itex]\pi[/itex]/4.. I was worried asking it haha it has been a long time since I have had to work with complex numbers.. Another thing I have just come across is the region {z : |z - 3 + i| < 4} Does this mean that z0 = (-3,i), and the radius = 4?
     
  6. Mar 4, 2012 #5

    tiny-tim

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    no, the centre is 3 - i
     
  7. Mar 4, 2012 #6
    With the first part from your first reply I said radius = a because I am trying to sketch the particular region covered by these angles or is that wrong?
     
  8. Mar 4, 2012 #7

    tiny-tim

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    I still don't understand this at all.

    What is a, and what has the radius to do with anything? :confused:
     
  9. Mar 4, 2012 #8
    Well I am not sure I just took it as an assuption.. See if I try and sketch this region I draw both these angles taking them anti-clockwise from the origin, which leaves a region in the 1st and 2nd quadrants and I am just confused as I thought I was suppose to be left with a closed region but is this not the case? I am sorry if this still makes no sense it is hard to explain a drawing in words. :/ So currently I have a line in the direction of [itex]\pi[/itex] going anti-clockwise from (0,0) and then another line in the direction of [itex]\pi[/itex]/4 from (0,0) Is that how the region is suppose to look?
     
  10. Mar 4, 2012 #9

    tiny-tim

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    Yup. There's no restriction on |z|, so the region goes to infinity.

    Goodnight! :zzz:​
     
  11. Mar 4, 2012 #10
    Oh okay thanks so much for all your help and sticking with me through all my confusion!! I appreciate it :D
     
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