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.
(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.
Puzzle Solution
