1987 AHSME Problems/Problem 28

Revision as of 23:03, 25 June 2016 by Lucasxia01 (talk | contribs) (Problem)

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 magnitude of these 4 roots are 1 as stated 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, 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
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