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Sketch the region of the complex plane

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Sketch the region of the complex plane specified by:

    |z - 4 + 3i| ≤ 5


    2. Relevant equations



    3. The attempt at a solution
    I have tried re-writing the modulus as √[(z)^2 (- 4)^2 + (3i)^2] and from this I have managed to arrive at z ≤ 3√2

    But not sure if I needed to do this or how I would take it from here in terms of sketching this
     
  2. jcsd
  3. Oct 31, 2013 #2

    tiny-tim

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    Hi SteveDC! :smile:
    |z - (4 - 3i)| ≤ 5 ? :wink:
     
  4. Oct 31, 2013 #3
    Sorry, I might need a bigger hint then this! I still don't really understand
     
  5. Oct 31, 2013 #4

    tiny-tim

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    how would you draw |z| ≤ 5 ? :wink:
     
  6. Oct 31, 2013 #5
    As a line along the real axis stretching to a point that is less than or equal to 5?
     
  7. Oct 31, 2013 #6

    pasmith

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    That's not how you compute [itex]|z - 4 + 3i|[/itex]. Recall that if [itex]w = a + ib[/itex] then [itex]|w|^2 = a^2 + b^2[/itex].

    Set [itex]z = x + iy[/itex] and see what happens.
     
  8. Oct 31, 2013 #7

    HallsofIvy

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    In any set in which an absolute value is defined we can interpret |x- y| as the distance between x and y. In particular, in the complex plane, |z- a| is the distance between z and a. If [itex]|z- b|\le r[/itex], for z a variable, b a specific complex number, and r a real number, then z is any point on or inside the circle with center at b and radius r.

    (If z is a complex number, [itex]z\le 3\sqrt{2}[/itex] makes no sense. The complex numbers are not an "ordered field".)
     
  9. Oct 31, 2013 #8

    tiny-tim

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    (just got back :tongue2:)
    ahh … that's where your misunderstandning is …

    |z| ≤ 5 is a circle, the circle of all points whose distance from 0 is ≤ 5

    i] do you see why that is? (or do you need an explanation?)

    ii] now what does |z - i| ≤ 5 look like?
     
  10. Nov 1, 2013 #9
    Think I've got this now. ii] a circle round the midpoint at i, with radius less than or equal to 5, and z will lie on that radius.
     
  11. Nov 1, 2013 #10
    If by "midpoint" you mean "centered", then you are correct.
     
  12. Nov 1, 2013 #11
    Yep, thanks everyone
     
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