Consider the following polynomial equation with $q_4, q_3, q_2, q_1$ , and $q_0$ real constants:
$$q_4z^4 + iq_3z^3 + q_2z^2 + iq_1z + q_0 = 0$$
If $z = a + bi$ is a solution to this equation, where a and b are real constants and $i^2 = -1$ , which of the following must also be a solution?
$(\text{A}) -a-bi$ $(\text{B}) a-bi $ $(\text{C}) -a+bi $ $(\text{D}) b+ai $ $(\text{E}) \text{None of these}$
I chose B (I thought this was the obvious answer because of conjugates). However, it's C. Does the $i$ in front of $q_3$ and $q_1$ affect the answer? I noticed that in B, it has the $bi$ negated, and in C, the $a$ is negated.
Consider the change of variable $t=iz$. The given equation can be written as $$q_4t^4-q_3t^3-q_2t^2+q_1t+q_0=0,$$ in which case we can apply the standard result that roots of polynomials with real coefficients appear as conjugate pairs. So if $t_1=-b+ai$ is a solution, then $t_2=-b-ai$ is also a solution. But, in terms of $z$, these solutions are $z_1=a+bi$ and $z_2=-a+bi,$ as the answer provides.