It is obvious for $a=0$. Now consider $a\ne0$. We can find all the roots of form $x=e^{i\theta}$. Divide the polynomial with $x^n$, we have equation $2\cos(n\theta)+2a\sum_{k=1}^{n-1}\cos(k\theta)+a=0$. Multiply by $\sin(\theta/2)$, $\sin((n+\frac12)\theta)-\sin((n-\frac12)\theta)+a(\sin((n-\frac12)\theta)-\sin(\theta/2))+a\sin(\theta/2)=0$, or $\sin((n+\frac12)\theta)-(a-1)\sin((n-\frac12)\theta)=0$, or $\sin(n\theta)\cos(\theta/2)(2-a)+a\cos(n\theta)\sin(\theta/2)=0$, or $\tan(n\theta)(1-\frac2a)=\tan(\frac{\theta}2)$. We can easily see the equation has one root on $(\frac{2k-1}{2n}\pi,\frac{2k+1}{2n}\pi)$ for $k=1,\cdots,n-1$. Paring with $-\theta$, we actually proved there are $n-1$ pairs conjugate roots to the polynomial of form $e^{i\theta}$.