Properties of a an entire function: Solution verification I would like to answer the following question

My solution:
(i) Since $f$ is entire it follows that $f$ is holomorphic and hence $f$ is analytic. Since $f$ is analytic on the whole $\mathbb{C}$ it follows that $f$ is analytic on the origin. Therefore we can write $f(z)= \sum a_n z^n $. If $f(z)/z \rightarrow 0$ as $|z| \rightarrow \infty$ then
$$\lim_{|z|\rightarrow \infty}\frac{\sum a_n z^n}{z}=\lim_{|z|\rightarrow \infty} \frac{a_0}{z} + a_1 + \sum_{n=2}^{\infty}a_n z^{n-1}=\lim_{|z|\rightarrow \infty}a_1 + \sum_{n=2}^{\infty}a_n z^{n-1}=0 $$
This must be true for all $z$ outside of the disc $D(0,N)$ (where $N$ is sufficiently large).  Since the function we are taking a limit of is analytic then by analytic continuation it follows that $a_1 + \sum_{n=2}^{\infty} a_nz^{n-1}=0$ for all $z$.  From this we get that
$$f(z)=a_0$$
and hence $f$ is constant.
(ii) Since $f$ is analytic it follows that if $f'(z)\neq 0$ then $f$ has an analytic isomorphism (by the inverse mapping theorem). The existence of the analytic isomorphism implies that $f$ is surjective and hence $f$ maps the whole of $\mathbb{C}$ to $\mathbb{C}$.  Now suppose that the condition in (ii) true. Since $f$ is subjective we can pick $z_0$ s.t. $f(z_0)=b$ and hence $|f(z_0)-f(z_0|=0$ which is a contradiction (could have also used the continuity of analytic functions here so that we have $|f(w=z_0+\delta)-f(z_0)|<\epsilon)$. Therefore the assumption that $f'(z)\neq 0$ must be false. Hence $f$ is constant.
My question: Is this correct?
 A: Part $1$ is fine, it is proved correctly.
For part $2$, we use Liouville's theorem on the function $\frac 1{f(z)-b}$, which is an entire function because the denominator is non-zero (and division, shifting etc. preserve entirety if valid), and also is a bounded function since $$\left|\frac 1{f(z)-b}\right|= \frac 1{|f(z)-b|} < \frac 1{\epsilon}$$
for all $z$. Hence $\frac 1{f(z)-b}$ is constant, therefore $f$ is constant.

The inverse mapping theorem tells you this : if $f'(z) \neq 0$ at a point , then there exist neighbourhoods $U,V \subset \mathbb C$ with $z \in U$ and $f(z) \in V$ such that $f: U \to V$ is analytic and a bijection.
The problem with what you want to say is two-fold. First of all, what if $f'(z)=0$ at some point? Note that if $f'$ is entirely $0$, only then is $f$ constant, but if $f$ is not constant it is still possible for $f'$ to have zeros. So at these points you can't create such a map.
Next, even if the function has non-zero derivative at all points, it still doesn't mean it is surjective. This is because, even if for each point you take a neighbourhood around it for which the isomorphism exists, the $V$s put together don't give you the entire of $\mathbb C$. Take the exponential function as an example of this phenomena.
Here is a nice result on what the image of an entire map must be.

Little Picard Theorem : Every nonconstant analytic function omits at most one complex number from its range. So at most one number cannot be the image of a non-constant entire map. The exponential map omits $0$, for example, but nothing else.

