1996 AIME Problems/Problem 11
Problem
Let be the product of the roots of that have a positive imaginary part, and suppose that , where and . Find .
Solution
Solution 1
Thus , or (see cis). Discarding the roots with negative imaginary parts (leaving us with ), we are left with ; their product is .
Solution 2
Let the fifth roots of unity, except for . Then , and since both sides have the fifth roots of unity as roots, we have . Long division quickly gives the other factor to be . The solution follows as above.
Solution 3
Divide through by . We get the equation . Let . Then . Our equation is then , with solutions . For , we get . For , we get (using exponential form of ). For , we get . The ones with positive imaginary parts are ones where , so we have .
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |