1982 AHSME Problems/Problem 27

Problem

Suppose $z=a+bi$ is a solution of the polynomial equation $c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0,$ where $c_0, c_1, c_2, c_3, a,$ and $b$ are real constants and $i^2=-1.$ Which of the following must also be a solution?

$\textbf{(A)}\ -a-bi\qquad \textbf{(B)}\ a-bi\qquad \textbf{(C)}\ -a+bi\qquad \textbf{(D)}\ b+ai \qquad \textbf{(E)}\ \text{none of these}$

Solution

Let $z=wi,$ so the given polynomial equation becomes $\begin{align*} c_4(wi)^4+ic_3(wi)^3+c_2(wi)^2+ic_1(wi)+c_0&=0 \\ c_4w^4+c_3w^3-c_2w^2-c_1w+c_0&=0. &&(\bigstar) \end{align*}$ Note that $(\bigstar)$ is a polynomial equation in $w$ with real coefficients.

We are given that $w=\frac{a+bi}{i}=b-ai$ is a solution to $(\bigstar).$ By the Complex Conjugate Root Theorem, we conclude that $w=b+ai$ must also be a solution to $(\bigstar),$ from which $z=(b+ai)i=\boxed{\textbf{(C)}\ -a+bi}$ must also be a solution to the given polynomial equation.

~MRENTHUSIASM

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1982 AHSME Problems/Problem 27

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