Determine all analytic $f$, wherefor $|f(z)| \leq C(|z|^ {3/2} + |z-1|^{-3/2})$ on $\mathbb{C}\backslash\{1\}$ for some $C>0$. Determine all analytic $f$, wherefor $|f(z)| \leq C(|z|^{3/2} + |z-1|^{-3/2})$ on $\mathbb{C}\backslash\{1\}$ for some $C>0$.
In the assignment f needs to have the following property as well:
$f(0)=f'(0)=0, f''(0) = 1$
i tried using the using an theorem of louiville, which states that:
if $f$ is holomorphic on $\mathbb{C}$ we say if $|f(z)|\leq M|z|^m$ for some $M<\infty$ and $m>0$, then $deg(f) \leq m$. 
The main problem is that i have two terms on the right hand side, on top of that i don't see how to use the extra properties of f.
I got the feeling i need to use an other theorem, any hints or theorems that can help me in this case?
Kees
 A: It follows from $|f(z)| \leq C(|z|^{3/2} + |z-1|^{3/2})$ that $f$
is bounded in a neighbourhood of $z=1$ and therefore has a
removable singularity at $z=1$. So $f$ can be extended to an
entire function.
For $|z|>1$ we have $|z-1| \le |z| + 1 < 2|z|$ and therefore
$$
|f(z)| \leq C(|z|^{3/2} + |z-1|^{3/2}) \le C(1+2^{3/2})|z|^{3/2} \le 5 C \, |z|^{3/2}
$$
It follows that $f$ is a polynomial of degree at most $1$. 

Update after question edit:
It follows from $|f(z)| \leq C(|z|^{3/2} + |z-1|^{-3/2})$ that $f$
has a pole of order at most one at $z=1$:
$$
 f(z) = \frac{a}{z-1} + g(z)
$$
where $a \in \mathbb C$ and $g$ is an entire function.
Then for $|z|>2$ we have $  |z-1|  > 1$ and therefore
$$
 |g(z)| \le |f(z)| + |a| < C( |z|^{3/2} + 1) + |a| \le (2C + |a|)|z|^{3/2} \, .
$$
It follows that $g$ is a polynomial of degree at most one,
and therefore
$$
 f(z) = \frac{a}{z-1} + b + c z \, .
$$
The values for $a, b, c$ can now be determined by your initial
conditions $f(0)=f'(0)=0, f''(0) = 1$.
