What are the possible poles of the meromorphic function $G$? 

Consider the function $$F(z)=\int_{1}^{2} \frac {1} {(x-z)^2}dx,\ \mathrm {Im} (z) > 0.$$ Then there is a meromorphic function $G(z)$ on $\Bbb C$ which agrees with $F(z)$ when $\mathrm {Im} (z) > 0$, such that 
$1.$ $\ 1, \infty$ are poles of $G(z)$.
$2.$ $\ 0,1,\infty$ are poles of $G(z)$.
$3.$ $\ 1,2$ are poles of $G(z)$.
$4.$ $\ 1,2$ are simple poles of $G(z)$.


If we evaluate $F$ we get $F(z)=\frac {1} {(1-z)(2-z)},\ \mathrm {Im} (z) > 0$. So if we define a function $G$ by $G(z)=\frac {1} {(1-z)(2-z)}$ then $G$ is analytic everywhere except for the simple poles at $1$ and $2$ and agrees with $F$ on $\mathrm {Im} (z) > 0$ since $\{z \in \Bbb C : \mathrm {Im} (z) > 0 \} \subset  \Bbb C \setminus \{1,2 \}$. So clearly $(3)$ and $(4)$ are the correct options. I think $(1)$ and $(2)$ are not correct but what is the reason behind that? I got stuck here. Any help in this regard will be appreciated.
Thank you very much.

Two meromorphic functions on $\mathbb C$ that agree on a nonempty open set are identically equal.  So there is only one possible $G$, and it satisfies (3) and (4) but not (1) and (2). 
