Constructing a rational function given its divisor My question concerns divisors of rational functions. I have tried searching up similar questions, such as Constructing meromorphic function from a given divisor. , but that question constructs only one function. Does there exists other functions with that same divisor?
My question: Construct all rational functions $f$ (quotients of polynomials) in $\mathbb{C}$ such that $(f)=6[0]-4[5]$, where $(f)$ is the divisor of the function $f$.
My answer/attempt: From my understanding of divisors, there seems to be exactly one function, namely, $f(z)=\frac{z^{6}}{(z-5)^{4}}$.
Q1. Are there any other functions with this divisor? If not, why?
Q2. If we restrict ourselves to divisors of rational functions only, then if we are given a divisor of a $\Delta$ on $\mathbb{C}$, does there exists a unique rational function $f$ with $(f)=\Delta$?
Any help/comments are appreciated.
 A: The answer to your question is yes, but one must invoke some machinery.
On a smooth projective curve, all functions have $(f)$ a degree zero divisor. By adding a point at infinity we see that if $\Delta$ is a divisor on $\mathbb{A}^1_\mathbb{C}$, then for any $f \in \mathbb{C}(z)$ we have that $(f)$ is equal to $\Delta$ if and only if when extended to $\mathbb{P}^1$ we have $(f) = \Delta - \deg(\Delta)[\infty]$.
By the Riemann-Roch Theorem, since $\mathbb{P}^1_\mathbb{C}$ has genus $0$, we have that $\dim_{\mathbb{C}} H^0(\mathbb{P}^1_\mathbb{C}, \mathcal{O}(D)) = \deg(D) + 1$ (whenever $\deg(D) \geq -1$). Now take the divisor $D = -(\Delta - \deg(\Delta)[\infty])$, which has degree $0$. Then $\dim_{\mathbb{C}} H^0(\mathbb{P}_\mathbb{C}^1, \mathcal{O}(D)) = 1$.
We have then proved that there is a $1$-dimensional $\mathbb{C}$ vector space of functions $f \in \mathbb{C}(z)$ such that $(f) \geq -D$, but since $D$ has degree $0$, equality must hold. Thus (up to multiplication by a scalar) there is a unique function $f \in \mathbb{C}(z)$ such that the divisor of $f$ is $\Delta$ on $\mathbb{A}^1_\mathbb{C}$.
