the Extended Liouville's Theorem with upper bound of $|f(z)|$ as follows Let $f$ be an entire function. If there exists a positive real constant $K$ such that $|f(z)| \le K(|z|^2 + |z|)$ for all $z \in \Bbb C$, then find a formula for $f(z)$.
Attempt:
Define a function $g(z) = \frac{f(z)}{z^2+z}$ for all complex numbers $z \notin \{0,1\}$. Then, $g$ is bounded (by $1$), it is holomorphic in $\Bbb C \setminus \{0,1\}$, and have two removable singularities at $z=-1$ and $z=0$. Hence, $g$ can be extended to a function $G$ which is holomorphic in all of $\Bbb C$, and coincides with $g$ on $\Bbb C \setminus \{0,1\}$. But, I stuck in determine the function $G$. If such $G$ can be determined, by Liouville's theorem, it follows that $g$ is constant, and hence, $f(z) = C(z^2 + z)$ for all $z \in \Bbb C$.
How to approach it?
Thanks in advanced.
 A: We must have $f(0)=0$, so $f(z)=zg(z)$ for some entire function $g$. For this $g$ we have $|g(z)|\leq K(|z|+1)$. Consider the function $$h(z)=\frac{g(z)-g(0)}{z}.$$ $h$ is entire and moreover $$|h(z)|\leq \left|\frac{g(z)-g(0)}{z}\right|\leq\frac{|g(z)|+|g(0)|}{|z|}\leq \frac{K(|z|+1)+K}{|z|}=K+\frac{2K}{|z|}.$$ This implies that $\lim_{z\to\infty}|h(z)|\leq K$. Now since $|h(z)|\leq K+1$ on $|z|>R$ for some $R$ and $h(z)$ is bounded on $|z|\leq R$ since it is compact, we have that $h$ is bounded on all of $\mathbb{C}$ and it is therefore constant.
Then $$\frac{g(z)-g(0)}{z}=C$$ and $g(z)=Cz+g(0)$. This means that $g(z)$ is of the form $az+b$. Now $|b|=|g(0)|\leq K$ and $|a|=\lim_{z\to\infty}|h(z)|\leq K.$ We now get that $f$ must be of the form $az^2+bz$ where $|a|\leq K$ and $ |b|\leq K$.
Moreover if $f$ is any function of the form $az^2+bz$ where $a$ and $b$ are as above then it satisfies $|f(z)|=|az^2+bz|\leq |a||z|^2+|b||z|\leq K(|z|^2+|z|)$.
A: $g(z) = \frac{f(z)}{z^2+z}$ is not necessarily bounded in $\Bbb C \setminus \{0,1\}$ and need not have a removable singularity at $z=-1$. A simple counterexample is $f(z) = z^2-z$.
But $|f(z)| \le K(|z|^2 + |z|)$ implies that $f(0) = 0$, and that $f$ is a polynomial of degree at most two, see for example Entire function bounded by a polynomial is a polynomial. It follows that $f$ is of the form
$$
f(z) = a z + b z^2
$$
with constants $a, b \in \Bbb C$. Conversely, every such function satisfies the given growth restriction.
A: Since $f$ is an entire function, it must have a Maclaurin series representation, namely
\begin{equation*}
f(z) = c_0 + c_1 z + c_2 z^2 + c_3 z^3 + \ldots
\end{equation*}
whenever $|z| < \infty$, where
\begin{equation*}
c_n = \frac{f^{(n)} (0)}{n!} = \frac{1}{2\pi i} \int_C \frac{f(w)}{w^{n+1}} \ dw,
\end{equation*}
where $C$ is a circle (in the positive direction) of radius $R>0$ centered at $0$ and $n=0,1,2,\ldots$.
the Goal: Show that for an arbitrary integer $p \ge 1$ and a large enough $r > R$, we have $c_{2+p} = 0$.
Proof of the Goal: Let $p \ge 1$ be an arbitrary integer and $r>R$ be a large number. Notice that for a contour (in the positive sense) $C_1: |w| = r$,
\begin{align*}
|c_{2+p}| &= \left| \frac{1}{2\pi i} \int_{C_1} \frac{f(w)}{w^{p+3}} \ dw \right| \\
&\le \frac{1}{2\pi} \int_{C_1} \frac{|f(w)|}{|w|^{p+3}} \ dw \\
&\le \frac{1}{2\pi} \int_{C_1} \frac{K(|w|^2 + |w|)}{|w|^{p+3}} \ dw \\
&= \frac{K}{2\pi} \cdot 2\pi \cdot \left(\frac{1}{r^{p+1}} + \frac{1}{r^{p+2}} \right) \\
&= K \left(\frac{1}{r^{p+1}} + \frac{1}{r^{p+2}} \right).
\end{align*}
Now, letting $r \to \infty$, we have $|c_{2+p}| \le 0$. Hence, $c_{2+p} = 0$, as desired. Q.E.D.
Thus, $f(z)=c_0 + c_1z+c_2z^2$ whenever $|z|< \infty$. Now, from $|f(z)| \le K(|z|^2 + |z|)$, we have $f(0)=0$ forcing $c_0 = 0$. Therefore, $f(z) = az^2 + bz$ for some constants $a,b \in \Bbb C$.
