Question

Suppose $z=a+bi$ is a solution of the polynomial equation $\begin{aligned} c_4z^4+ic_3z^3+c_2z^2\\ {}+ic_1z+c_0=0, \end{aligned}$ where $c_0,$ $c_1,$ $c_2,$ $c_3,$ $c_4,$ $a,$ and $b$ are real constants and $i^2=-1.$ Which one of the following must also be a solution?

$-a-bi$

$a-bi$

$-a+bi$

$b+ai$

none of these

Answer: C

Concepts:complex number symmetry (algebra)polynomial

Difficulty rating: 2090

Small Hint:

Conjugate the entire equation

Big Hint:

Compare the conjugated equation at $\bar z$ with the original polynomial evaluated at $-\bar z$

Solution:

Conjugating the equation changes each $i$ to $-i$ and $z$ to $\bar z.$ Because the odd-powered terms also change sign when the input is negated, this conjugated equation is precisely the original polynomial evaluated at $-\bar z.$ Thus $-\bar z=-a+bi$ must be a root.

Therefore, the correct answer is C.

1982 AMC 12 Problem 27

Answer: C

Difficulty rating: 2090

Small Hint:

Big Hint:

Solution:

Problem 27 in Other Years