I just picked up Wilhelm Schlag's book A Course in Complex Analysis and Riemann Surfaces, and I've been stuck on a Exercise 1.3:
Let $P(z)=\sum_{j=0}^{n}a_jz^j$ be a polynomial of degree $n\geq 1$ with all roots inside the unit circle $|z|<1$. Define $P^*(z)=z^n\overline{P}(z^{-1})$ where $\overline{P}(z)=\sum_{j=0}^{n}\overline{a}_jz^j$. Show that all roots of $$P(z) + P^*(z)=0$$ lie on the unit circle $|z|=1$.
Here's what I've worked out so far. By the fundamental theorem of algebra we can write $$P(z)=a_n\prod_{j=1}^{n}(z-z_j),$$ where $z_j$ are the roots of $P$. By definition of $P^*$ we see that $P^*(z)=\overline{a}_n\prod_{j=1}^{n}(1-z\overline{z}_j)$, which immediately yields (again by the fundamental theorem of algebra) $$P^*(z)=\overline{a}_0\prod_{j=1}^{n}(z-\overline{z}_j^{-1}).$$ So we see that the roots of $P^*$ are just the inversions of the roots of $P$ around the unit circle. There is something a little funny about this: if zero is a root of $P$ then there is no associated root of $P^*$ (this makes some sense since polynomials can't vanish at $\infty$), and so the degree of the polynomial drops by one in this case?
Now we can compute $(P+P^*)^*=(P+P^*)$ and so by the above arguments all of the roots of $P+P^*$ are symmetric (in the sense of inversions) around the unit circle. Now here's where I get stuck... I'm not sure how to use the assumption that the roots of $P$ (denoted above by $z_j$) lie inside the unit disk to show that the roots of $P+P^*$ must lie exactly on the unit disk...
Hints on where to go from here are much more appreciated than complete solutions!