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Complex Numbers Inequality

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine the region in the complex plane described by |z-2i| < |z+ i|

    2. Relevant equations

    z= x+ iy
    |z|= (x2 + y2)1/2


    3. The attempt at a solution

    |z-2i| < |z+ i|

    |z-2i|/|z+ i| < 1

    |z-2i| = [(x-2i)2 + y2]1/2
    |z+ i| = [(x+i)2 + y2]1/2

    [(x-2i)2 + y2]1/2
    --------------- < 1
    [(x+i)2 + y2]1/2


    [(x-2i)2 + y2]1/2*[(x+i)2 - y2]1/2
    ---------------------------------- < 1
    [(x+i)2 + y2]1/2*[(x+i)2 - y2]1/2



    Am I on the right track of solving this so far? If so how do I proceed to the next step? If not what part did I do wrong? Any feedback is appreciated!
     
  2. jcsd
  3. Sep 12, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    |z-i2|^2 is NOT (x-2i)^2 + y^2. Think about why not.

    RGV
     
  4. Sep 12, 2011 #3

    dynamicsolo

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    Homework Helper

    Trying to compute the inequality in Cartesian coordinates is an efficient way to make yourself crazy.

    It may be more helpful to use a geometrical interpretation of the equation first in the Argand diagram. Keep in mind that | z - z0 | is the "length" of a vector from the point representing z0 to the point representing z . The equation | z - 2i | = | z + i | then describes the curve in the Argand diagram of points equidistant from 2i and -i . What is that curve like? The inequality then represents the set of points closer to 2i than to -i . Where is that region?
     
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