Complex analysis - Prove a holomorphic function has at least one high-order pole

Question

I feel very confused about the following problem. Much appreciate if someone can help me.

Let f(z) be holomorphic everywhere on the complex plane apart from n points $a_1$, $a_2$, ... $a_n$ (with each $a_i\neq 0$), where it has simple holes. Consider another function $g(z)$ which is holomorphic everywhere on the complex plane except at the same n points $a_1$, ... $a_n$, where it has isolated singularities. Assume that $f(0)\neq g(0)$ and ${\rm Res}_{z=a_i}\{f(z)\} = {\rm Res}_{z=a_i}\{g(z)\}$ for all $i$. And $\lim_{z\rightarrow \infty} f(z) = \lim_{z\rightarrow \infty} g(z)$ and the limit exists. Prove that at least one of the singularities of $g(z)$ is a pole of order greater than 1, or an essential singularity.

Answer 1

Clearly, $$ h(z)=f(z)-\sum_{k=1}^n \frac{r_k}{z-a_k} $$ has the same limit at infinity with $f$, and it is equal to $$ c=\lim_{z\to\infty} f(z). $$ Hence $h$ is bounded and entire and hence constant. Thus $$ f(z)=c+\sum_{k=1}^n \frac{r_k}{z-a_k}. $$ If all the singularities of $g$ were poles of first order, then $g$ would have the same expression and be identical to $f$. But they are not identical.

Answer 2

$f-g$ is analytic on $\Bbb{C}$ minus finitely many points where it has zero residues, also $f-g$ is continuous at $\infty$ thus it is bounded for $|z|> r$.

If all those points are removable singularities then $f-g$ is entire and bounded thus it is constant, and $f(\infty)-g(\infty)=0$ means $f=g$,

otherwise $f-g$ has at least one pole of order $\ge 2$ or an essential singularity.