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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 Puzzle Solution Comment 4 of 4 |
(In reply to Answer by K Sengupta)

From equation (b), we have:

z^6 - z^5 - z + 1 = 0
or, (z-1)^2*(z^4 + z^3 + z^2 + z + 1) = 0

Thus, z = 1, or:

z^4 + z^3 + z^2 + z + 1 = 0

But, z=1 in condition (c) gives :

1 - r + 1 = 0, so that r=2, which contradicts condition (a).

Accordingly,

z^4 + z^3 + z^2 + z + 1 = 0
or, z^4 + z^3 + z + 1 = - z^2
or, z^2 + z^(-2) + z + 1/z = -1 ........(#)

But, condition (c), we have:

z^2 + 1 = r*z

or, r = z + 1/z
or, r^2 - 2 = z^2 + z^(-2)

Substituting this back to (*), we have:

r^2 - 2 + r = -1
or, r^2 + r - 1 = 0
or, r = (-1 +/- V5)/2

But, r = (-1 - v5)/2 is negative and thus contradicts condition (a).

Consequently, r = (v5 - 1)/2 =  0.618034 (correct to six places) is the only possible solution.


  Posted by K Sengupta on 2008-07-21 04:26:11
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