smc

Question

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?

Accepted Answer

Let z=wi, so the given polynomial equation becomes 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. &&() align* Note that () is a polynomial equation in w with real coefficients. We are given that w=a+bii=b-ai is a solution to (). By the Complex Conjugate Root Theorem, we conclude that w=b+ai must also be a solution to (), from which z=(b+ai)i=-a+bi must also be a solution to the given polynomial equation.