Show that $f (z) = cz^n$ for some $c ∈ \Bbb C$ and $n ∈ N.$ 
Suppose $f ∈ H (\Bbb C)$, $f (0) = 0,$ and
{$z : |f (z)| < M$} is connected $∀ M > 0$
Show that $f (z) = cz^n$
for some $c ∈ \Bbb C$ and $n ∈ N.$

So I think it follows from the fact that $f(z)$ is bounded and entire for which it has a constant. Any hint or which theorem I can use to show this? I really tried to think of something but I only come with this that $f(z)$ is bounded but how exactly to show that it is $f (z) = cz^n$?
 A: Let us ignore the trivial case that $f$ is identically zero.
Since $f(0) = 0$ we can choose a radius $R > 0$ such that $f(z)  \ne 0$ for $0 < |z| \le R$, and a real number $M$ with
$$
  0 < M < \min \{ |f(z)| : |z| = R \} \, .
$$
The level set $\{ z : |f (z)| < M \}$ contains the origin and does not intersect the circle $|z| = R$. Since it is connected, it follows that this level set is completely contained in the disk $\{ |z| < R \}$. In other words, we have
$$ \tag{$*$}
 |f(z)| \ge M \text{ for } |z| \ge R \, .
$$
This implies that

*

*$f$ has no zeros except at the origin, and

*$f$ does not have an essential singularity at $\infty$ (e.g. by the Casorati-Weierstrass theorem), and is therefore $f$  a polynomial.

Finally, a polynomial which is non-zero except at the origin is necessarily of the form $f(z) = cz^n$.
Alternative argument, as suggested by Greg Martin: From $(*)$ we see that $f$ has no zeros except at the origin. It follows that
$$
 f(z) = \frac{z^n}{g(z)}
$$
with $g \in H (\Bbb C)$, $n$ is the multiplicity of the zero of $f$ at the origin. Then $(*)$ gives
$$
 |g(z)| \le \frac{R^n}{M} \text{ for } |z| \ge R \, ,
$$
which implies that $g$ is bounded, and therefore constant (by Liouville's theorem).
