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|4+z| - |4-z| = 6, prove |4+z|^2 - |4-z|^2 >= 48

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Where z is a complex number:

    |4+z| - |4-z| = 6

    Prove that |4+z|^2 - |4-z|^2 >= 48

    2. The attempt at a solution

    |4+z| = 6 + |4-z|

    |4+z|^2 = 36 + |4-z|^2 + 12|4-z|

    |4+z|^2 - |4-z|^2 = 36 + 12|4-z|

    From here, I figure if I can prove that |4-z| >= 1, I can prove that |4+z|^2 - |4-z|^2 >=48

    Any pointers as to how I can do this? Or am I approaching this from the wrong angle?
     
  2. jcsd
  3. Jan 30, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    You could look at it geometrically. |4-z| is the distance from 4 to z. If |4-z|<1 then |z+4| (the distance from z to -4) is going to be greater than 7. So the difference can't be 6.
     
  4. Jan 30, 2012 #3
    |4-z| = |z-4|, so my geometric model is of a point z in the midline of a "horizontal" line of length 8 and the two legs to the origin differ in length by 6. This defines a curve of possible values for z, which crosses the real axis at ...
     
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