I haven't been around complex analysis at some time now and I need it for Riemann surfaces and I a bit confused about something. I know that if I have a meromorphic function $f$ with order $k$ at $p$ and a meromorphic $g$ with order $n$ at $p$ I will have that $ord_p(f+g)\geq min\{ord_p f, ord_pg\}$ and smaller then their max, but can we be a bit more specific, by this I mean is it possible to have all cases in the previous example, that is I can have that $f+g$ will have order $k$ with $k$ varying between the maximum and minimum of the previous valies? I remember that there was a theorem that we could find the order of poles a function taking the limits of $(z-z_0)f(z)$ and finding the first $k$ such that this limit is zero, and with this my intuition would say no, but again I am not that familiar with complex analysis so any help is aprecciated.
New edit: I am interested in the case that suppose I have $3$ points and $ord_{p_1}f=k_1$ $ord_{p_2}f=k_2$ and $ord_{p_3}f=m$ and $ord_{p_1}(g)=k_1$ $ord_{p_2}(f)=n$ $ord_{p_3}(g)=k_3$ where these are all negative numbers such that $n>k_2$ and $m>k_3$. Is it possible that from $f_1$ and $f_2$ I can construct a meromorphic function such that the order at $p_i$ will be $k_i$, It's just like you said we dont know what is happening when the functions have the same order so I cant really sum them.
I am trying to do this because I am trying to do a Riemann surfaces exercise that is that if we have a base-point free linear system $|D|$ we will have that for any finite number of points $\{p_1,…,p_k\}$ we have that there exists $E\in |D|$ such that $p_i$ is not in the support of $E$ and that is basically proving the existence of such meromorphic function I belive.
Thanks in advance!!
Best Answer
If $f$ and $g$ have poles of orders $k$ and $n$ at $p$ and $k < n$, then $f+g$ has a pole of order $n$ at $p$. This is obvious if you consider the Laurent series around $p$. If $k=n$, all you can say is the order $\le k$: note that you could even have $f+g=0$.