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Sketching inequalities involving complex numbers

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Sketch all complex numbers z which satisfy the given condition:


    2. Relevant equations



    3. The attempt at a solution

    First I find the boundary between the regions where the inequality holds and does not hold by replacing the inequality sign with an equality sign. Then I substitute z=a+bi into the equation and solve for b:




    So the boundary cuts through the origin and has a slope of -1.

    To find which side of the boundary the equality holds, I can plug random values from different sides of the boundary, in this case z=-1-i and z=1+i.

    Doing this, I get [itex]\sqrt{-3}\geq\sqrt{3}[/itex] and [itex]\sqrt{1}\geq\sqrt{-1}[/itex]

    What do I do from here?
  2. jcsd
  3. Jun 7, 2012 #2
    This is incorrect. [itex]|z| = \sqrt{a^2+b^2}[/itex]

    You can also solve this without using a and b, in a rather simple way. This how I would try, by drawing the graph. [itex]|z-i|[/itex] represents all circles(vectors) with center at (0,1) and [itex]|z-1|[/itex] represents circles with center (1,0). The perpendicular bisector of the line joining these two points gives the condition when [itex]|z-i|=|z-1|[/itex]. So, which side of the curve would you sketch z, such that [itex]|z-i|>|z-1|[/itex]?

    Try drawing a graph, and it should become really simple to figure out :smile:
  4. Jun 7, 2012 #3
    Ah, I thought i was included in the formula. I re-did the entire thing with the correct form, and I got it now. Thanks! :)
  5. Jun 7, 2012 #4
    Graphically, modulus/absolute value represents the magnitude of distance between the point and origin. Keeping that in mind, its easy to see why it comes out to be [itex]\sqrt{a^2+b^2}[/itex]
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