Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs/Proof 2

Theorem
Let $P$ be the polynomial equation:
 * $a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = 0$

such that $a_n \ne 0$ and such that $a_0, a_1, \ldots, a_n$ are all real.

Let $p + q i$ be a root of $P$.

Then $p - q i$ is also a root of $P$.

Proof
Let $p + q i$ be expressed in exponential form as $r e^{i \theta}$.

As $r e^{i \theta}$ satisfies $P$, it follows that:
 * $a_n r^n e^{n i \theta} + a_{n-1} r^{n-1} e^{\left({n - 1}\right) i \theta} + \cdots + a_1 r e^{i \theta} + a_0 = 0$

Taking the conjugate of both sides:
 * $a_n r^n e^{-n i \theta} + a_{n-1} r^{n-1} e^{-\left({n - 1}\right) i \theta} + \cdots + a_1 r e^{-i \theta} + a_0 = 0$

it follows that $p - q i$ is also a root of $P$.

If any of the $a_k$ are complex, then the conjugate of $P$ is not the same polynomial as $P$.

Therefore the result holds only for real coefficients.