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

  • Thread starter matt.lmx
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  • #1
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Homework Statement


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?
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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|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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