Question 1207465:  if |(z)/(|conj(z)|) -conj(z) | = 1 + | z | , z \[Element] C , then (z imaginary) prove that ? 
 Found 2 solutions by  Edwin McCravy, ikleyn: Answer by Edwin McCravy(20064)      (Show Source):  Answer by ikleyn(52903)      (Show Source): 
You can  put this solution on YOUR website! . 
if |(z)/(|conj(z)|) -conj(z) | = 1 + | z | , z \[Element] C , then (z imaginary) prove that ? 
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The equation in the post is 
   |      z                |
   | ----------  - conj(z) |  = 1 + |z|    (1)
   |  |conj(z)|            |
Vertical lines denote absolute value, or the modulus of complex numbers.
|conj(z)| is the same as |z|.
        z                          z
So, ----------   is the same as  ----,  which is the unit vector, co-directed with z.
     |conj(z)|                    |z|
So, equation (1) tells us that a unit vector, co-directed with z, MINUS conj(z) 
has the absolute value (the modulus) of 1+|z|.    (2)
Taking into account the rule of adding/subtracting complex numbers as vectors,
we can interpret the statement (2) THIS WAY :
    we have a triangle with three sides: side  "a"  is the unit vector, co-directed with z;
                                         side  "b"  is the vector -conj(z);
                                         side  "c"  has the length 1+|z|.
Keeping in mind a triangle inequality, we can conclude that this triangle is degenerated
and its sides "a", "b" and "c" all are collinear.  Namely,
    +-------------------------------------+
    |    -conj(z) is co-directed with z   |
    +-------------------------------------+
which may happen IF and ONLY IF  z is a purely imaginary complex number.
It is the statement to prove, and at this point, the solution is COMPLETE
and the problem' statement is proven.
 
Solved.
 
 
 
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Really nice problem.
 
 
 
Re-phrasing the famous American writer O'Henry,
 
 
        the blind begin to walk and the dumb to see 
 
        when they receive so brilliant solutions.
 
 
 
 
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