1. ## Proofs about an analytic function

Let $f(z)$ be an analytic function in the open disk $D(0,R)$ which satisfies $|f(z)| \leq |z|^n$ for some $n \in \mathbb{N}^*$ and for all $z \in D(0,R)$.
1)Prove that $f$ has a zero in $z=0$ or order $\geq n$.
2)Prove that $\frac{f(z)}{z^n}$ has a removable singularity at $z=0$.
3)Assume that $f$ is entire and that $|f(z)| \leq |z|^n$ for some $n \in \mathbb{N}^*$ and for all $z \in \mathbb{C}$. Show that $f(z)=c z^n$ with $c \in \mathbb{C}$.
Attempt:1) Let $g(z)=|z|^n$. Then we have $g(0)=0$. Since $|f(z)| \leq g(z)$, it follows that $f(0)=0$. Does this also mean that $|f^{(n)}(z)|\leq n! |z|$? If so, then I think I've proven part 1).
Part 2): I think they want me to prove that $\lim _{z \to 0} z\frac{f(z)}{z^n}=l$ where $l \in \mathbb{C}$.
Using the fact that $\frac{|f(z)|}{|z|^n} \leq 1$, I get that $-z \leq \frac{f(z)}{z^{n-1}} \leq z$ 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.

2. For the first one argue with the series expansion at 0. Could $g(z)=\frac{f(z)}{z^n}$ be bounded around 0 if there was a coefficient $a_k\neq 0$ with $0\leq k\leq n$ in the Taylor series for f?

For the second one, if g is bounded near 0, could it have a pole or essential singularity there?

For the third apply Liouville to g.

3. I'm still clueless with 1) and 3).
With 2), $-1 \leq g(z) \leq 1$. I just saw a theorem in my class notes which states that if g is analytic in $D(0,R)-z_0$, then $z_0$ is a pole of g if and only if $\lim _{z \to z_0} f(z)=\infty$. So clearly there is no pole at z=0 for g. I must still show that there isn't an essential singularity. Writing $f(z)=\sum _{m=0}^{+\infty} a_m z^m$ since it's analytic, it's obvious that dividing it by $|z|^n$ won't make appear an infinity of negative powers of z. Therefore since g(0) isn't defined, it can only be a removable singularity.

4. Originally Posted by arbolis
With 2), $-1 \leq g(z) \leq 1$.
Be careful, you made the same mistake in your first post: We are dealing with complex numbers, which aren't ordered, so the inequality above is meaningless.

For the first one note that $|g(z)| \leq 1$ and apply the same reasoning as in (2) and use the fact that $f(z)$ has a zero of order n at 0 iff $f(z)=z^nh(z)$ where h is analytic and $h(0)\neq 0$.

For the third one, using the second one you get that $g(z)$ has an entire extension whose modulus is bounded by 1