Let $C$ be $$C=\{[X,Y,Z] \in \mathbb{P}^2_\mathbb{C} \mid X^4-Y^4+Z^4=0\}$$ and $p=[0,i,1]$. Compute $l(np)$ for $n\geq0$.
My attempt:
For $n\geq5$ we have $l(np)=n-2$ by the riemann roch theorem.
Since the curve has degree 4 the tangent in $p$, which is $4iY+4Z=0$ and has degree 1, intersects the curve in $4p$. So we have that $4p$ is canonical and thus $l(4p)=3$.
This also gives us $l(0p)=1$.
I'm at a loss on how to proceed in the remaining cases. I've tried to find a base of $L(1p)$, $L(2p)$ and $L(3p)$ by hand but it's proving to be a somewhat difficult process (and one that can't really be extended to other situations elegantly).
Is there any general way to deal with these situations? I've tried looking up what's written in this answer but while I understood some of it I didn't actually manage to apply it to the problem in exam here.
After consulting with my professor the following seems to be the solution. First of all a disclaimer, I don't really grasp the concepts behind this procedure so if anyone has a source or knows more about this argument please comment and let me know.
Solution:
Let $L$ be the tangent in $p$. We have that $L\cap C=4p$. Now let us consider two other lines:
$R$, which intersects $C$ in $p$ but is not tangent, and $S$, which doesn't intersect the curve in $p$.
Now let's consider $L/L,\mbox{ }R/L$ and $S/L$. The first is a constant function and is present in all the $L(np)$. The second is a function with a pole of third order in $p$, and thus $3p$ is in $L(np)$ for $n\geq3$ and finally $S/L$ has a pole of order 4 and appears from $L(4p)$ onwards.
So we have: $$l(0p)=l(1p)=l(2p)=1$$ $$l(3p)=2$$ $$l(4p)=3$$ $$l(5p)=3$$