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Real and Complex (Posted on 2007-01-23) Difficulty: 3 of 5
Let r and z be real and complex numbers respectively, such that

(a) 0 < r < 1
(b) z^6 - z^5 - z + 1 = 0
(c) z^2 - rz + 1 = 0

Find the value of r.

See The Solution Submitted by Dennis    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 1 of 4

 z^6 - z^5 - z + 1 = 0
   ==> (z - 1)(z^5 - 1) = 0
   ==> z = 1 or z^5 = 1.
   ==> z = cos(72k) + i sin(72k)
           for k = 0, +-1, or +-2
 z^2 - rz + 1 = 0
   ==> r = z + 1/z = z + conjugate(z) 
         = 2 Real(z) = 2 cos(72k)
 0 < r < 1 
   ==> r = 2 cos(72) ~= 0.618034
  
 

  Posted by Bractals on 2007-01-23 15:32:50
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