Difference between revisions of "1987 AHSME Problems/Problem 28"

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==Solution==
 
==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>
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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 12:44, 31 March 2018

Problem

Let $a, b, c, d$ be real numbers. Suppose that all the roots of $z^4+az^3+bz^2+cz+d=0$ are complex numbers lying on a circle in the complex plane centered at $0+0i$ and having radius $1$. The sum of the reciprocals of the roots is necessarily

$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ -a \qquad \textbf{(E)}\ -b$

Solution

Let's denote the roots of the polynomial as $z_1, z_2, z_3, z_4$. We know that the magnitudes of these 4 roots are 1 as given in the problem statement. Therefore, we have $z_1 \overline{z_1}, z_2 \overline{z_2}, z_3 \overline{z_3}, z_4 \overline{z_4} = 1$. We want to find $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \frac{1}{z_4}$ which is $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4}$. 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, $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4} = z_1 + z_2 + z_3 + z_4$. However, by Vieta's, $z_1 + z_2 + z_3 + z_4 = -a$. Thus, the answer is $\boxed{\textbf{(D) } -a }$. --$\text{lucasxia01}$

See also

1987 AHSME (ProblemsAnswer KeyResources)
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

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