Find all holomorphic functions on $\mathbb{C}$, except for some singularities, such that $|f(z)|\leq C(|z|^{3/2}+|z-1|^{-3/2}), z\in\mathbb{C}-\{1\}$ First I wrote the Laurent series of $f(z)$ around $z=1$:
$$
f(z)=\sum_{n=-\infty}^{-1}c_n(z-1)^n+\sum_{n=0}^{\infty}c_n(z-1)^n.
$$
Now if $|z|$ becomes very large, the first sum with the negatives powers becomes very small. So for some $\varepsilon >0$ we have 
$$
\left|\sum_{n=0}^{\infty}c_n(z-1)^n\right|\leq C|z|^{3/2}+\varepsilon\leq(\varepsilon '+C)|z|^{3/2}.$$
 Now I want to use Louiville's theorem, but I don't know how that works here exactly. Can I say that all terms after $n=2$ are $0$? I need some help on this on this exercise. Thanks in advance.
 A: Let $g(z)=\sum_{n=0}^{\infty}c_n(z-1)^n$. Once you have $|g(z)|\le C|z|^{3/2}$ for large $z$, an application of Cauchy integral formula for derivatives implies $g^{(n)}(1)=0$ for $n\ge 2$. This is shown here. Therefore, $g(z) = c_0+c_1(z-1)$.  
Concerning the first sum, $h(z)=\sum_{n=-\infty}^{-1}c_n(z-1)^n$, you can observe that $(z-1)^2 h(z)$ tends to $0$ as $z\to 1$. Therefore, the singularity of $(z-1)^2 h(z)$ is removable: extending this function by $0$ at $z=1$, we get a holomorphic function. It follows that $h(z) = c_{-1}(z-1)^{-1}$. 
Conclusion: 
$$f(z) = c_{-1}(z-1)^{-1}+ c_0+c_1(z-1)$$
A: First observe that 
$$|f(z) (z-1)^2 | \leq C(|z|^{3/2}|z-1|^2+|z-1|^{1/2}), z\in\mathbb{C}-\{1\}$$
therefore $f(z)(z-1)^2$ is bounded in a neighborhood of each of its singularities. This implies that all the singularities of $f(z)(z-1)^2$ are removable.
Thus, there exists an entire function $g(z)$ such that $g(z)=f(z) (z-1)^2$ excepting at the singularities of $f$. Since $g$ is continuous, we have
$$|g(z)| \leq C(|z|^{3/2}|z-1|^2+|z-1|^{1/2}), z\in\mathbb{C}-\{1\}$$
Now, taking the limit when $z \to 1$ we get $g(1)=0$ and therefore $g(z)=(z-1) h(z)$ for an entire function $h$.
This Yields
$$f(z)(z-1)=h(z)$$
for some entire function $h$.
Now write 
$$h(z)=a+bz+cz^2j(z)$$
where $j$ is entire.
The inequality becomes 
$$|(z-1) (a+bz+cz^2j(z)) | \leq C(|z|^{3/2}|z-1|+|z-1|^{-1/2}), z\in\mathbb{C}-\{1\})$$
Divide both sides by $(z-1)z^2$ and deduce that $j(z)$ is a bounded entire function..Moreover, taking the limit at infinity, you get that the constant must be $0$.
