Prove $|f(z)| < 5 + |z|^{\frac{1}{3}} \implies f$ is constant 
$f$ is an entire function and $|f(z)| < 5 + |z|^{\frac{1}{3}}$ for all $z \in \Bbb C$. Prove $f$ is a constant.

So I want to bound my function in order to use Liouville's theorem, but in order to bound my function, I'm not sure how to do this. I looked at Cauchy's Estimates that says if $f$ is holomorphic on $B(a, R) = \{z : |z - a| < R\}$ and $|f(z)| \leq M$ for some number $M > 0$ and for all $z \in B(a, R)$, then $|f^n (a)| \leq \frac{n! M}{R^n}$ ($f^n (a)$ is the $n$th derivative of $f$ at $a$).
However, in our given inequality we have a variable and not a number. How can I deal with this?
 A: For all $ z \in B(0, R)$ we have
$$
  |f(z)| < 5 + |z|^{1/3} < 5 + R^{1/3}  \, .
$$
Then the Cauchy estimate gives for all $R > 0$
$$
  |f^{(n)}(0)| \le \frac {n! (5 + R^{1/3}) }{R^n}
$$
With $R \to \infty$ you can conclude that $f^{(n)}(0) = 0$ for all
$n \ge 1$.
So the power series of $f$ has only the constant term.
A: Let $f(z)=a + zg(z)$. Then $|\frac a z + g(z)|= |\frac {f(z)} z| \leq \frac 5 {|z|} + \frac 1 {|z|^{\frac 2 3}}$. For $|z|\to \infty$ this gives $\lim |g(z)| =0$, which shows that $\exists R>0$ such that $g(z)$ is bounded for $|z|>R$. Since $g$ is holomorphic, it will also be bounded for $|z|\leq R$, which shows that $g$ is bounded on the whole $\mathbb{C}$. Then, by Liouville's theorem, $g$ must be constant, so $f=a+zg$ ($g$ constant now!). Now, divide by $|z|^{\frac 2 3}$ and you'll get $|\frac a {|z|^{\frac 2 3}} + \frac z {|z|^{\frac 2 3}} g| \leq \frac 5 {|z|^{\frac 2 3}} + \frac 1 {|z|^{\frac 1 3}}$. For $|z|\to \infty$ this gives $\infty \leq 0$, unless $g=0$ and then $f=a$, a constant.
