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(20054)   (Show Source):
Answer by ikleyn(52756)  (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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