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

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. 
