How to show that $p(1)\ \text{is real} \iff \ p(-1)\ \text{is real}$

Question

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 $$p(1)\in\mathbb{R} \;\;\Longleftrightarrow\;\; p(-1)\in\mathbb{R} $$

Answer 1

Note that (assuming $p(-1)\ne 0$ to begin with $$ \frac{p(1)}{p(-1)}=\prod_{j=1}^{2k}\frac{1-w_j}{-1-w_j}$$ where the $w_j$ run over the complex roots (with multiplicity). For a single factor, $$\frac{1-w}{-1-w}=-\frac{(1-w)(1+\bar w)}{|1+w|^2}=\frac{|w|^2-1+(w-\bar w)}{|1+w|^2}. $$ As we are given that $|w|=1$ for all roots, this is the purely imaginary number $\frac{2\operatorname{im} w}{|1+w|^2}i$. The product of an even number of imaginaries is real.

Answer 2

We have that

$$p(z)=\prod_{r=1}^{2k}\left(z-e^{i\theta_r}\right))$$

then assuming $p(1)\neq 0$

$$p(1)=\prod_{r=1}^{2k}(1-e^{i\theta_r})=\prod_{r=1}^{2k}(1-e^{-i\theta_r})=\prod_{r=1}^{2k}\frac{1-e^{i\theta_r}}{e^{i\theta_r}} \iff \prod_{r=1}^{2k}\frac{1}{e^{i\theta_r}}=1$$

$$p(-1)=\prod_{r=1}^{2k}(-1-e^{i\theta_r})=\prod_{r=1}^{2k}(-1-e^{-i\theta_r}) \iff \prod_{r=1}^{2k}\frac{1}{e^{i\theta_r}}=1$$