Let $f(z)=p(z)/q(z)$ where $p(z),q(z)$ are polynomials without common roots and with $deg(q)\ne deg(p)$, $\alpha\in \mathbb{C}-\{0\}$.
Let $n=\max\{\deg(p),\deg(q)\}$. Prove that $f(z)=\alpha$ has exactly $n$ solutions, counted with multiplicity.
How may I proceed with the proof?
Any help will be appreciated.
(Assume $\deg p \neq \deg q$) With the fundamental theorem of algebra: $f(z) = \alpha$ becomes $$\alpha q(z) - p(z) = 0$$ $\alpha q(z) - p(z)$ is a polynomial of degree $\max\{n,m\}$, whence we have that many solutions.
Without FTA it can be done with Rouche's theorem: find $R$ large enough such that in a disk of radius $R$ centered at $0$, $$|\alpha q(z) - p(z)| \leq |z^{\max\{n,m\}}|$$ The trouble with this approach is that one may use the same argument to prove the fundamental theorem of algebra, so in a backwards way we are still using it. It can also be done using the argument principle, but we have the same issue as above if we're trying to avoid FTA.