Let $\displaystyle f(z)$ be an analytic function in the open disk $\displaystyle D(0,R)$ which satisfies $\displaystyle |f(z)| \leq |z|^n$ for some $\displaystyle n \in \mathbb{N}^*$ and for all $\displaystyle z \in D(0,R)$.

1)Prove that $\displaystyle f$ has a zero in $\displaystyle z=0$ or order $\displaystyle \geq n$.

2)Prove that $\displaystyle \frac{f(z)}{z^n}$ has a removable singularity at $\displaystyle z=0$.

3)Assume that $\displaystyle f$ is entire and that $\displaystyle |f(z)| \leq |z|^n$ for some $\displaystyle n \in \mathbb{N}^*$ and for all $\displaystyle z \in \mathbb{C}$. Show that $\displaystyle f(z)=c z^n$ with $\displaystyle c \in \mathbb{C}$.

Attempt:1) Let $\displaystyle g(z)=|z|^n$. Then we have $\displaystyle g(0)=0$. Since $\displaystyle |f(z)| \leq g(z)$, it follows that $\displaystyle f(0)=0$. Does this also mean that $\displaystyle |f^{(n)}(z)|\leq n! |z|$? If so, then I think I've proven part 1).

Part 2): I think they want me to prove that $\displaystyle \lim _{z \to 0} z\frac{f(z)}{z^n}=l$ where $\displaystyle l \in \mathbb{C}$.

Using the fact that $\displaystyle \frac{|f(z)|}{|z|^n} \leq 1$, I get that $\displaystyle -z \leq \frac{f(z)}{z^{n-1}} \leq z$ which tend to 0 as z tends to 0 because of the sandwich theorem. Proved. (I don't think my proof is valid).

3)I'm not sure.