$\lim_{\vert z\vert\to\infty}\frac{zf'(z)}{f(z)}=n\in\mathbb{N}$ implies $f$ is a polynomial Let $f$ be an entire function, meaning $$f:\mathbb{C}\to\mathbb{C}$$ is holomorphic. If $f\not\equiv0$ and $$\lim_{\vert z\vert\to\infty}\frac{zf'(z)}{f(z)}=n\in\mathbb{N}_0$$ then $f$ have to be a polynomial of degree $n$.
I was able to proof that $f$ can only have finitely many roots, using
$$\frac{zf'(z)}{f(z)}=\frac{z}{z-z_1}+\frac{z}{z-z_2}+...$$
where $z_n$ are the roots. Taking the limit this series diverges if there were infintely many roots. This also already proofed, that $f$ has exactly $n$ roots. Nonetheless doing this is only valid, if $f$ even has any roots.
If there are no roots, $f$ can not be a non-constant polynomial. So it's either constant or it is a transcendental entire function. So proving that $f$ can not be an transcendental entire function would imply the case what happens when $f$ has no roots.
I would appreciate any hints how to prove, that $f$ is a polynomial.
 A: $$
 g(z) = \frac{zf'(z)}{f(z)}
$$
is meromorphic in $\Bbb C$, with a simple pole at each point $z \ne 0$ where $f$ has a zero. Since
$$
 \lim_{|z| \to \infty} g(z) = n \,,
$$
$g$ has only finitely many poles, so that $f$ has only finitely many zeros.
So we can write
$$
 f(z) = P(z) e^{h(z)}
$$
with a polynomial $P$ and an entire function $h$. (This covers also the  case that $f$ has no zeros at all, in that case $P$ is a constant polynomial.)
Then
$$
 \frac{zf'(z)}{f(z)} = \frac{zP'(z)}{P(z)} + z h'(z) \, .
$$
Now every polynomial satisfies
$$
\lim_{|z| \to \infty}\frac{zP'(z)}{P(z)} = \deg P
$$
so that
$$
 \lim_{|z| \to \infty} zh'(z) = n - \deg P \, .
$$
It follows that $zh'(z)$ is a bounded entire function, and therefore constant:
$$
 zh'(z) = n - \deg P \, .
$$
Setting $z=0$ gives $n = \deg P$ and consequently, $h'(z) = 0$.
So $h$ is constant, and $f(z) = P(z)e^h$ is a polynomial of degree $n$.
Remarks:

*

*As pointed out by Kavi Rama Murthy in the comments, it suffices to require that $L = \lim_{|z| \to \infty}\frac{zf'(z)}{f(z)}$ exists (as a real number). The proof still works and shows that $f(z) = P(z)$ is a polynomial, and $L = \deg P$ is in fact a non-negative integer.


*If $f$ is meromorphic function in $\Bbb C$ and $L = \lim_{|z| \to \infty}\frac{zf'(z)}{f(z)}$ exists then the same method shows that $f(z) = P(z)/Q(z) $ is rational and $L = \deg P - \deg Q$ is an integer.
