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

Proof
Let $\alpha = p + q i$.

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

As $\alpha = r e^{i \theta}$ satisfies $\map f \alpha = 0$, it follows that:
 * $a_n r^n e^{n i \theta} + a_{n - 1} r^{n - 1} e^{\paren {n - 1} i \theta} + \dotsb + a_1 r e^{i \theta} + a_0 = 0$

Taking the conjugate of both sides:
 * $a_n r^n e^{-n i \theta} + a_{n - 1} r^{n - 1} e^{-\paren {n - 1} i \theta} + \dotsb + a_1 r e^{-i \theta} + a_0 = 0$

it follows that $\overline \alpha = p - q i$ is also a root of $f$.

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

Therefore the result holds only for real coefficients.