complex analysis - Prove that when $z = p$ is a solution of $az^3+bz^2+cz+d=0$, $z=-p^*$ is also a solution

Given that $z = p$ is a solution of the equation:

$$az^3+bz^2+cz+d=0$$

where $a$ and $c$ are real constants while $b$ and $d$ are purely imaginary constants.
Show algebraically that $x = -p^*$ (negative conjugate) is another solution.

I tried to solve this question by letting $p = x + iy$, so $-p^* = -(x-iy) = -x + iy$.

So I tried to bring connections between $x + iy$ and $-x + iy$ by substitute them into the equation.

When I substitute x + iy, the equation becomes

$$a(x+iy)(x^2 - y^2 + 2xyi) + b(x^2 - y^2 + 2xyi) + c(x + iy) + d = 0$$

However, I failed to proceed with this question. By substituting $-x+iy$ into the equation, I cannot show that it is also a solution.

Is there a simpler method to solve this question?

Answer

Given the original polynomial:

$$az^3 + bz^2 + cz + d = 0$$

for some complex root $z$, we can take the complex conjugate of both sides:

$$\bar{a} \bar{z}^3 + \bar{b} \bar{z}^2 + \bar{c} \bar{z} + \bar{d} = 0$$

Here I've used just that complex conjugation preserves complex arithmetic (and that zero is its own complex conjugate).

Now use the fact that $a,c$ are real constants, while $b,d$ are purely imaginary:

$$a \bar{z}^3 - b \bar{z}^2 + c \bar{z} - d = 0$$

Comparing these coefficients to the original ones, we see only the even terms have changed sign. Throwing in an extra times $-1$, we have:

$$a (-\bar{z})^3 + b (-\bar{z})^2 + c (-\bar{z}) + d = 0$$

This shows, since we are back to the original coefficients now, that $-\bar{z}$ is a root whenever $z$ is a root.