Poles and Singularity Let $f(z)$ be an analytic function in the whole complex plane apart from simple poles at $z_1,...,z_m$. Moreover $f(1/z)$ has a simple pole at $0$. Show that $f$ is a ratio of two polynomials. 
To start this, we know $f(z)=\frac{A_1}{z-z_1}+...+\frac{A_m}{z-z_m}+h(z)$ where $h(z)$ is in the form $\sum_{k=1}^\infty a_nz^n$. Since $f(1/z)$ has a simple pole at zero. Then $h(z)$ must be polynomial. So $f(z)$ is a ratio of polynomials. Is this true and clear? Thanks
 A: Yes, it is true that if $h(1/z)$ has a simple pole at $z=0$, then $h(z)$ must be a polynomial but why? I think you need more argumentation. I like better to proceed in a slightly different way:
Consider the function $F: \mathbb{C} \to \mathbb{C}$ given by
$$
F(z) = f(z) \prod_{j=1}^m (z-z_j)
$$
Clearly $F$ is an entire function, since we are "removing" the poles of $f$. Since $f(1/s)$ has a simple pole at $s=0$, then also $F(1/s)$ has it, therefore it's Laurent series in the annulus $\{ 0< |s|<r \}$ for some $r>0$ is
$$
F(1/s) = \sum_{n\boldsymbol{= -}1}^{\infty} a_n s^n 
$$
Hence $s\mapsto sF(1/s)$ is analytic in $\{ 0< |s|<r \}$, thus there exist  $M>0$ such that $|sF(1/s)|\leq M$ if $s \in \{ 0< |s|<r \}$. Putting $z=1/s$ and $R=1/r$, we get
$$
|F(z) | \leq M |z|\ \text{ for $|z|>R$} \tag{1}
$$
But since $F$ is entire, line $(1)$ together with Cauchy estimates gives that $F$ must be a polynomial with degree at most 1. Therefore, indeed $f$ is a ratio of polynomials since 
$$
f(z) = \frac{F(z)}{ \prod_{j=1}^m (z-z_j)}
$$
