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Complex Number question

  1. May 19, 2013 #1
    1. The problem statement, all variables and given/known data

    The locus of z satisfying the inequality (z + 2 i) / (2 z + i) < 1 where z = (x + i y)

    2. Relevant equations

    none

    3. The attempt at a solution
    After putting the value of z = (x + i y) , I tried to rationalize it but now I am stuck. can somebody explain how to solve it. The correct answer is x2 + y2 < 1

    Thank you in advance.
     
    Last edited: May 19, 2013
  2. jcsd
  3. May 19, 2013 #2

    SammyS

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    You can't use inequalities directly with complex numbers.

    Also, you need to use parentheses in your expressions to make them represent what you actually mean.

    I assume you mean to be working with ( z + 2 i )/( 2 z + i)

    z + 2 i / 2 z + i literally means ##\ \displaystyle z+\left(2\left(\frac{i}{2}\right)z\right)+i\ .##
     
  4. May 19, 2013 #3

    haruspex

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    Do you mean |(z + 2 i) /( 2 z + i)| < 1?
     
  5. May 19, 2013 #4
    yes that is what I meant
     
  6. May 20, 2013 #5

    haruspex

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    Ok. Given an expression like 1/(x+iy), can you see how to get it into the form u+iv?
     
  7. May 20, 2013 #6
    Are you meaning to say that I have to rationalize it ?
     
  8. May 20, 2013 #7

    haruspex

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    Yes.
     
  9. May 20, 2013 #8
    read before you post

    Well, if you had only carefully read my first post. I did rationalize it but I am stuck at that.
     
  10. May 20, 2013 #9

    Dick

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    You don't need to rationalize it. Use |a/b|=|a|/|b|. So your inequality becomes |z+2i|<|2z+i|, right? Just express both sides in terms of x and y.
     
  11. May 20, 2013 #10
    :eek: Why didn't I think of it before ??? But anyways thanks for the hint :smile:
     
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