2004 AMC 12A Problems/Problem 23

Problem

$P(x) = c_{2004}x^{2004} + c_{2003}x^{2003} + ... + c_1x + c_0$

has real coefficients with $c_{2004}\not = 0$ and $2004$ distinct complex zeroes $z_k = a_k + b_ki$ , $1\leq k\leq 2004$ with $a_k$ and $b_k$ real, $a_1 = b_1 = 0$ , and

$\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}.$

Which of the following quantities can be a nonzero number?

$\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}$

Solution 1

We have to evaluate the answer choices and use process of elimination:

$\mathrm{(A)}$ : We are given that $a_1 = b_1 = 0$ , so $z_1 = 0$ . If one of the roots is zero, then $P(0) = c_0 = 0$ .
$\mathrm{(B)}$ : By Vieta's formulas, we know that $-\frac{c_{2003}}{c_{2004}}$ is the sum of all of the roots of $P(x)$ . Since that is real, $\sum_{k = 1}^{2004}{b_k}=0=\sum_{k = 1}^{2004}{a_k}$ , and $\frac{c_{2003}}{c_{2004}}=0$ , so $c_{2003}=0$ .
$\mathrm{(C)}$ : All of the coefficients are real. For sake of contradiction suppose none of $b_{2\ldots 2004}$ are zero. Then for each complex root $z_i$ , its complex conjugate $\overline{z_i} = a_i - b_ik$ is also a root. So the roots should pair up, but we have an odd number of imaginary roots! (Remember that $b_1 = 0$ .) This gives us the contradiction, and therefore the product is equal to zero.
$\mathrm{(D)}$ : We are given that $\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}$ . Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.

There is, however, no reason to believe that $\boxed{\mathrm{E}}$ should be zero (in fact, that quantity is $P(1)$ , and there is no evidence that $1$ is a root of $P(x)$ ).

Solution 2(cheap method using answer choices)

Rule out answer choices $A, B, D$ as done in solution 1. Assume for the sake of contradiction that $\mathrm{(E)}$ is $0$ . Then $P(1)=0$ , so there is sum $m\neq 1$ such that $z_m=1+0i$ , which implies at least one $b_i$ , $i\neq 1$ is $0$ SO if $\sum_{k=1}^{2004}c_k=0$ , then $\prod_{i=12}^{2004}b_i=0$ . So we have that $\sum_{k=1}^{2004}c_k\neq 0$ , because then that would rule out all the answer choices. Hence $\boxed{\mathrm{E}}$ must be nonzero, so the answer is $\mathrm{(E)}$ . -vsamc

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2004 AMC 12A Problems/Problem 23

Contents

Problem

Solution 1

Solution 2(cheap method using answer choices)

See also