How to show that $p(1)\ \text{is real} \iff \ p(-1)\ \text{is real}$ I have been working on this problem and I cannot figure it out! I spent hours of time on it with no use. Can anybody help? The question is:
Suppose $p(x)$ is a polynomial with complex coefficients and even degree($n=2k$). All zeros of $p$ are non-real and with length equal to $1$. prove $$p(1)\in\mathbb{R} \;\;\Longleftrightarrow\;\; p(-1)\in\mathbb{R} $$

We have that
$$p(z)=\prod_{r=1}^{2k}\left(z-e^{i\theta_r}\right))$$
then assuming $p(1)\neq 0$
$$p(1)=\prod_{r=1}^{2k}(1-e^{i\theta_r})=\prod_{r=1}^{2k}(1-e^{-i\theta_r})=\prod_{r=1}^{2k}\frac{1-e^{i\theta_r}}{e^{i\theta_r}} \iff \prod_{r=1}^{2k}\frac{1}{e^{i\theta_r}}=1$$
$$p(-1)=\prod_{r=1}^{2k}(-1-e^{i\theta_r})=\prod_{r=1}^{2k}(-1-e^{-i\theta_r}) \iff \prod_{r=1}^{2k}\frac{1}{e^{i\theta_r}}=1$$
