Removable singularities and an entire function

Question

Function $f(z)$ is an entire function such that $$|f(z)| \le |z^{n}|$$ for $z \in \mathbb{C}$ and some $n \in \mathbb{N}$.

Show that the singularities of the function $$\frac {f(z)}{z^{n}}$$ are removable. What can be implied about the function $f(z)$ if moreover $f(1) = i$? Draw a far-reaching conclusion.

My attempt: If the singularities of $\frac {f(z)}{z^{n}}$ are removable, it is entire (not sure, need help with the justification) and bounded, so constant from Liouville's theorem, the constant value of the funcion is $i$, hence $f(z)=iz^{n}$.

But what about the $n$ here, is it arbitrary? Could somebody help me prove the removability of the singularities and suggest if my attempt is going the right way?

Answer 1

Since $f(z)$ is an entire function, $g(z)=\frac{f(z)}{z^n}$ may only have a pole at the origin. If that was the case, then $\left|g(z)\right|$ would be unbounded in a neighbourhood of the origin, but that contradicts $\left|g(z)\right|\leq 1$ for any $z\neq 0$. It follows that $z=0$ is a removable singularity for $g(z)$ and $g(z)$ is an entire function. Since $g(z)$ is bounded, by Liouville's theorem it follows that $g(z)$ is constant, so $f(z)=C z^n$. If $f(1)=i$, it follows that $f(z)=i z^n$.

Answer 2

Consider $\frac{f(z)}{z^n}$. Since $f$ was entire, we just need to show that the singularity at $0$ is removable. Because of Reimann's theorem(read here https://en.wikipedia.org/wiki/Removable_singularity#Riemann.27s_theorem) this is equivalent to showing that

$$\lim_ {z \to 0}z\frac{f(z)}{z^n} = \lim_ {z \to 0}\frac{f(z)}{z^{n-1}} =0 $$ But this follows from your information that $|f(z)| \leq |z|^n$.

Rest of your proof was correct. As far as $n$ is concerned, it will be equal to your earlier $n$.(Prove this !)