Difference between revisions of "1987 AHSME Problems/Problem 28"
(Created page with "==Problem== Let <math>a, b, c, d</math> be real numbers. Suppose that all the roots of <math>z^4+az^3+bz^2+cz+d=0</math> are complex numbers lying on a circle in the complex pla...") |
Lucasxia01 (talk | contribs) (→Problem) |
||
Line 9: | Line 9: | ||
\textbf{(D)}\ -a \qquad | \textbf{(D)}\ -a \qquad | ||
\textbf{(E)}\ -b </math> | \textbf{(E)}\ -b </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Let's denote the roots of the polynomial as <math>z_1, z_2, z_3, z_4</math>. We know that the magnitude of these 4 roots are 1 as stated in the problem statement. Therefore, we have <math>z_1 \overline{z_1}, z_2 \overline{z_2}, z_3 \overline{z_3}, z_4 \overline{z_4} = 1</math>. We want to find <math>\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \frac{1}{z_4}</math> which is <math>\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4}</math>. 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, <math>\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4} = z_1 + z_2 + z_3 + z_4</math>. However, by Vieta's, <math>z_1 + z_2 + z_3 + z_4 = -a</math>. Thus, the answer is <math>\boxed{\textbf{(D) } -a }</math>. --<math>\text{lucasxia01}</math> | ||
== See also == | == See also == |
Revision as of 23:03, 25 June 2016
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.