Mock AIME 3 Pre 2005 Problems/Problem 4: Difference between revisions
mNo edit summary |
|||
| Line 2: | Line 2: | ||
<math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are [[complex number]]s such that | <math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are [[complex number]]s such that | ||
< | <cmath>\zeta_1+\zeta_2+\zeta_3=1</cmath> | ||
\zeta_1^2+\zeta_2^2+\zeta_3^2 | <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</cmath> | ||
\zeta_1^3+\zeta_2^3+\zeta_3^3 | <cmath>\zeta_1^3+\zeta_2^3+\zeta_3^3=7</cmath> | ||
Compute <math>\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}</math>. | Compute <math>\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}</math>. | ||
Revision as of 01:46, 19 March 2015
Problem
and 
are complex numbers such that
![\[\zeta_1+\zeta_2+\zeta_3=1\]](http://latex.artofproblemsolving.com/2/0/f/20f1cdf3667c7b392addc8b1658aef13e0a65918.png)
![\[\zeta_1^2+\zeta_2^2+\zeta_3^2=3\]](http://latex.artofproblemsolving.com/1/9/0/1905da989b086b38a25f8d01de7119ca8dc675d1.png)
![\[\zeta_1^3+\zeta_2^3+\zeta_3^3=7\]](http://latex.artofproblemsolving.com/f/c/a/fcaa6db09874632947ab41ccd40e1f90e3311b44.png)
Compute 
.
Solution
We let 
(the elementary symmetric sums). Then, we can rewrite the above equations as
from where it follows that 
. The third equation can be factored as
from where it follows that 
. Thus, applying Vieta's formulas backwards, and are the roots of the polynomial
Let 
(the power sums). Then from 
, we have the recursion 
. It follows that 
.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
