1987 AHSME Problems/Problem 28
Problem
Let be real numbers. Suppose that all the roots of are complex numbers lying on a circle in the complex plane centered at and having radius . The sum of the reciprocals of the roots is necessarily
Solution
Let's denote the roots of the polynomial as . We know that the magnitude of these 4 roots are 1 as stated in the problem statement. Therefore, we have . We want to find which is . Remember that these are the roots of polynomials. Whenever, complex numbers are the roots of a polynomial, we know they come in complex-conjugate pairs. Therefore, . However, by Vieta's, . Thus, the answer is . --
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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