Difference between revisions of "1987 AHSME Problems/Problem 28"
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\textbf{(D)}\ -a \qquad | \textbf{(D)}\ -a \qquad | ||
\textbf{(E)}\ -b </math> | \textbf{(E)}\ -b </math> | ||
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| + | ==Solution== | ||
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| + | Let's denote the roots of the polynomial as <math>z_1, z_2, z_3, z_4</math>. We know that the magnitudes of these 4 roots are 1 as given 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 with real coefficients, 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 == | ||
Latest revision as of 11:44, 31 March 2018
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 magnitudes of these 4 roots are 1 as given 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 with real coefficients, 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. 
