Removable singularity and Liouville's Theorem Please verify if I got either correct.
1a. Let $f$ be entire function such that $\sup_{\mathbb C} \left |\frac{f(z)}{z} \right | < \infty$. Show $z = 0$ is a removable singularity of $g(z) = \frac{f(z)}{z}$.
1b. Suppose $f$ and $g$ are entire functions such that $|f| \leq K|g|$, show that $f = cg$ for all $z \in \mathbb{C}$.
I wrote that
1a. Since $\sup |f(z)/z| < \infty$ then $\left |\frac{f(z)}{z} \right | < M$. Hence $\lim_{z \to 0}|zg(z)| = \lim_{z \to 0}|f(z)| \leq \lim_{z \to 0} M|z| =0.$ So passing the limit to both sides yields the result.
1b. I think this is just applying Liouville Theorem to $(f/g)$ never mind I found the answer. My answer is incomplete for 1b. Only need verification for 1a.
Thanks for reading.
 A: Technically, there is a problem with the question. More precisely, one should write $$\sup_{z\in{\mathbb C}\setminus\{0\}}\left|\frac{f(z)}z\right|<\infty,\quad (1)$$ as $g(z)=\frac{f(z)}{z}$ may be undefined at $z=0$.
1(a). Clearly (1) implies that $f(0)=0$, otherwise by continuity there exists $r>0$ such that $$|f(z)|\geq\frac 12|f(0)|\neq 0,\forall z~{\rm with~}|z|\leq r.$$ It then follows that $$\sup_{z\in{\mathbb C}\setminus\{0\}}\left|\frac{f(z)}z\right|\geq\frac 12|f(0)|\sup_{0<|z|\leq r}\frac 1{|z|}=\infty,$$ a contradiction.
Now since $f(z)$ is entire and $f(0)=0,$ one has $$\lim_{z\rightarrow 0}\frac{f(z)}z=\lim_{z\rightarrow 0}\frac{f(z)-f(0)}{z-0}=f’(0).$$ By Riemann’s Removable Singularity Theorem, $z=0$ is a removable singularity of $g(z)$, and $g(z)$ is analytic if $g(0)$ is defined as $f’(0).$
1(b). Define $h(z)=\frac{f(z)}{g(z)}$ for all $z\notin S:=\{z~|~g(z)=0\}.$ By a similar argument as in 1(a), any $z_0\in S$ is a removable singularity of $h(z),$ where one removes singularity by defining $h(z_0)=f^{(k)}(z_0)/g^{(k)}(z_0)$ (with $k$ the multiplicity of the zero $z_0$ for $g(z)$). Now $h(z)$ is entire and bounded, so by Liouville, one has $h(z)=c$ is a constant, hence $f(x)=cg(z),$ as required.
Remark. In 1(b), if $\sup_{{\mathbb C}\setminus S}|h(z)|$ is bounded, one can work out Riemann’s removable singularity theorem by power series expansion. If $z_0\in S$, then analogous to the proof in 1(a), one has $m\geq n$, where $m$ (resp. $n$) is the multiplicity of zero $z_0$ of $f(z)$ (resp. $g(z)$). Expanding in power series at $z_0$, one has $$f(z)=(z-z_0)^mf_1(z)=a_n(z-z_0)^n+\cdots+a_m(z-z_0)^m+\cdots$$ and $$g(z)=(z-z_0)^ng_1(z)=b_n(z-z_0)^n+\cdots,$$ where $$f_1(z_0)\neq 0,g_1(z_0)\neq 0,a_n=\frac{f^{(n)}(z_0)}{n!},b_n=\frac{g^{(n)}(z_0)}{n!}\neq 0.$$ (Note that $a_n=\cdots=a_{m-1}=0$ if $m>n$.)
After canceling common zeros at $z_0$, one sees that $\frac{f(z)}{g(z)}=(z-z_0)^{m-n}\frac {f_1(z)}{g_1(z)}$ has power series expansion at $z_0$ with constant term $$\frac{a_n}{b_n}=\frac{f^{(n)}(z_0)}{g^{(n)}(z_0)},$$ which is the value to be redefined for $h(z_0)$. (Note that $h(z_0)=0$ if $m>n$.)
A: As , $f(z)$ is entire, so, $g(z)=\frac{f(z)}{z} $ is analytic on $0 \lt |z-0| \lt \delta $ ,
Now, since $sup_{\mathbb{C}} |g(z)| \lt \infty $ , so, $g(z)$ must be bounded.
Hence, by Riemann's removable singularity theorem, $g(z)$ is either analytic or having removable singularity at $z=0$.
Analyticity is possible, if we redefine the function as ,  $F(z) =
\begin{cases}
\frac{f(z)}{z},  & \text{if $z \neq 0 $} \\
0, & \text{if $z = 0 $}
\end{cases}
$
Edit: Now as you have shown , $\lim_{z \to 0} |zg(z)| \le 0
 \implies |\lim_{z\to 0 } zg(z)| \le 0 
 \implies \lim_{z\to 0 } zg(z) = 0 $
( as, modulus function is continuous )
From these directly , you can conclude $g(z)=\frac{f(z)}{z} $ have removable singularity at $z=0$.
So, your answer now seems ok.
