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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 Excel solution | Comment 2 of 4 |
formulae:
      A                                                            B
1     theta                                                        rad
2     1.25663706143782                                             1.00000000000005
3
4     =B2^6*COS(6*A2)-B2^5*COS(5*A2)-B2*COS(A2)+1                  =B2^6*SIN(6*A2)-B2^5*SIN(5*A2)-B2*SIN(A2)
5
6     r=                                                           0.618033988735794
7
8     =B2^2*COS(2*A2)-B2*B6*COS(A2)+1                              =B2^2*SIN(2*A2)-B2*B6*SIN(A2)
values:
1     theta                                                        rad
2         1.256637061                                                            1
3
4        -9.19842E-12                                                 -6.34337E-12
5
6     r=                                                               0.618033989
7
8         3.15548E-12                                                  9.99989E-12

where A2 represents the theta portion of z expressed as polar form and B2 represents the radius.  These were each initially set arbitrarily to 1. Solver was used to set A4 and B4 (the rectangular Argand coordinates of the left side of (b)) to zero by changing A2 and B2.  Then solver was used a second time to set A8 and B8 (the rectangular coordinates of (c)) to zero by changing B6, the value for r being sought.

That value shows as 0.618033989, which amounts to 1/phi (to the accuracy shown) where phi is the golden mean.


  Posted by Charlie on 2007-01-23 16:19:15
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