TITLE: $P(z)-zQ(z)=0$ having no solution in $\mathbb{C}$
QUESTION [5 upvotes]: I am currently reading "Iteration of Rational Functions" by Alan F. Beardon and my Question is about the beginning of $§2.6.$ which deals with Fixed points. Given a rational function $R=P/Q$ where $P,Q$ are coprime polynomials. If $z_0 \neq \infty$ is fixed by $R$, then $Q(z_0)\neq 0$ and $P(z_0)=z_0Q(z_0).$ It is now stated that the fixed points of $R$ in $\mathbb{C}$ are the solutions of $P(z)-zQ(z)=0$.
After that it is noted that this need not have any solutions in $\mathbb{C}$ and the counter example given is $z \mapsto z+\frac{1}{z}$.
So far so good. My question is now that I do not see how to get this counterexample with the given solution.

REPLY [2 votes]: (Expanded from the posted comment.)
For no solutions to exist, $P(z)-zQ(z)$ must be a constant non-0 polynomial (otherwise if $\deg (P-zQ) \ge 1$ then it would have a root by the FTA, else if $P \equiv 0$ then $\forall z \in \mathbb{C}$ would be a root). Therefore no solutions exist iff $P(z) = z Q(z) + c$ for some $c \ne 0$. 
The given counterexample uses $Q(z)=z$ and $c=1\,$, so that $P(z)=z\cdot z+1=z^2+1$ and $R(z)=\cfrac{z^2+1}{z}=z+\cfrac{1}{z}\,$.