Range of any non-constant rational function $\textbf{Question}$. We know that the range of any nonconstant complex polynomial is $\mathbb{C}$ by the Fundamental Theorem of Algebra. What can you say about the range of any nonconstant rational function $f$?
$\textbf{Attempt / thoughts}$. Write $f=p/q$ and we try to solve $p(z)/q(z)=\lambda$ because this is possible if and only if $\lambda$ is in the range of $f$.
Now define $g(z)=p(z)-\lambda q(z)$. Solving the previous equation is equivalent to solving $g(z)=0$ except that we need to take care of the case where $q(z)=0$ (I think, I'm not sure). If $g(z)$ is a nonconstant complex polynomial then it will have a root and then we are done. So we consider the values of $\lambda$ such that $g(z)$ is a nonconstant complex polynomial... which I think is hard to determine.
Would really appreciate some help on how to proceed or advice on whether I'm on the right track at all.
 A: First, eliminate common factors between $p$ and $q$. This can be done using Euclid's algorithm for polynomials. This leaves us with two polynomials $P,Q$, still with $f=P/Q$.
Let $m = \deg{P}$, $n = \deg{Q}$. We call $\max{\{m,n\}}$ the degree of $f$, $\deg{f}$. 


*

*Suppose first that $m \neq n$,


*

*$P-\lambda Q$ is always nonconstant if $\lambda \neq 0$, because both polynomials can't be constant since they have different degrees (at least one of $m,n$ is $ \geq 1 $), and because they have different degrees, the terms cannot all cancel one another out. Hence $f(z) = \lambda$ has $\deg{f}$ solutions, counted with multiplicity.

*On the other hand, if $m \neq n$ and $\lambda = 0$, it is possible that $m=0$, in which case $P-\lambda Q = P$ is constant. Otherwise, the argument is the same, so $f(z)=0$ has $\deg{P}$ solutions.


*Now suppose that $m=n$ ($\neq 0$, since $f$ is nonconstant), and let the leading coefficients of $P$ and $Q$ be $a$ and $b$. Then 


*

*if $\lambda \neq a/b$, $P-\lambda Q$ is of degree $m$, since the leading terms don't cancel, so again $f(z)=\lambda$ has $\deg{f}$ solutions.

*if $\lambda = a/b$, the leading term and some number of non-leading terms cancel, but since $P/Q$ is nonconstant, $Q$ is not a multiple of $P$: indeed, by the division algorithm there is a polynomial $R \neq 0$ so that $ P = (a/b)Q+R $. Then $P-(a/b)Q=R$, so $f(z)=a/b$ has $\deg{R}$ solutions. This may be $0$: $P(z)=z+1$ and $Q(z)=z$, for example.
In conclusion:

Suppose that $f=P/Q$ where $P,Q$ are polynomials with no common factor. Then 
  
  
*
  
*If $m > n$, $f(z)=\lambda$ has $m$ solutions with multiplicity, whatever the value of $\lambda$.
  
*If $m<n$, $f(z)=\lambda$ has $n$ solutions with multiplicity if $\lambda \neq 0$, and $m$ solutions with multiplicity if $\lambda=0$.
  
*If $m=n$ and $P = \mu Q + R$ for a polynomial $R$ with $r=\deg{R}<m$, then $f(z)=\lambda$ has $m$ solutions with multiplicity if $\lambda \neq \mu$ and $r$ solutions with multiplicity if $\lambda=\mu$.
  

One could use the remainder $R$ in all cases to make the proofs a bit more uniform, but there's not a lot to be gained by doing so. On the other hand, it is essential that $P$ and $Q$ have no common factors, or we run into issues with roots of $P-\lambda Q$ not being roots of $P/Q-\lambda$.
