Showing that a function is constant using Liouville's theorem Let $α ∈ (0, 1)$ be a fixed exponent, let $C > 0$, and let
$f$ be an entire function satisfying $|f(z)| ≤ C|z|^α$
Show that $f$ is constant.
My attempt:
I'll show that using Liouville's theorem.
Let $z=x+iy$, $(x, y) \in \mathbb R^2$, then $|z| = {(x^2+y^2)}^{1/2}$ and so $|z|^α = {(x^2+y^2)}^{α/2}$.
Thus, $|f(z)|≤ C{(x^2+y^2)}^{α/2}$. 
And since $\mathbb C$ is endowed with $\mathbb R^2$, we can rewrite the above as $|f(x, y)|≤ {(x^2+y^2)}^{α/2}$, which shows that $f$ is continuous on $\mathbb C$ for all $z$. And we know that continuity implies boundness. So $f$ is also bounded. 
And since it is entire, by Liouville's, it's constant. 
Is my attempt correct? 
Any other answers? 
 A: Much like the proof of Liouville's Theorem, we use the Cauchy estimates
$$|\frac{f^{(n)}(0)}{n!}|\leq\frac{1}{2\pi}\int_\gamma\frac{|f(\zeta)|}{R^{n+1}}d\zeta$$
where $\gamma$ is a circle of radius $R$ centred at 0.
If $|f(\zeta)|\leq C|\zeta|^{\alpha}$, we get the following estimate
$$|\frac{f^{(n)}(0)}{n!}|\leq\frac{1}{2\pi}\int_\gamma CR^{-n-1+\alpha} d\zeta = CR^{ -n+\alpha}$$
which tends to 0 as $R$ tends to $\infty$ for $\alpha\in(0,1)$ and $n\geq 1$. Hence $ \frac{f^{(n)}(0)}{n!}=0$ and so, considering the Taylor series of $f$, we have $f(z)=f(0)$
A: Hint: $\frac{f(z)-f(0)}{z}$ is an entire function that goes to zero at infinity by hypothesis; what can you conclude?
A: Another way.
$|f(0)| \leq C|0|^\alpha = 0$, so $f(z)/z$ is also an entire function.  It satisfies,
$$  \left| \frac{f(z)}{z} \right| \leq C|z|^{\alpha - 1}  \text{.}  $$
$|z|^{\alpha -1}$ is a decreasing function on $|z| \geq 1$, with maximum $1$ attained on $|z| = 1$.  $f(z)/z$ is continuous on the compact set $|z| \leq 1$, so attains a maximum, $M$, on the closed unit disk.  Therefore, $f(z)/z$ is an entire function bounded by $\max \{C,M\}$ on $\Bbb{C}$ and by Liouville's theorem, $f(z) / z$ is a constant, $K$.
So we have $f(z) = Kz$ and we are given $|Kz| \leq C |z|^\alpha$.  Then $|K| \leq C|z|^{\alpha - 1}$ for all $z \neq 0$.  Taking the limit as $|z| \rightarrow \infty$, we find $K = 0$.  Therefore, not only is $f$ constant, it is the zero function.
A: Every entire function can be represented by a power series about any $a\in\Bbb C$
$$f(z) = \sum_{n\geqslant 0}c_n{(z-a)}^n$$
that converges for all $z\in\Bbb C$.
Pick $a=0$ and consider what Cauchy's Integral Formula tells you about $c_n$ for $n\geqslant 1$. (Hint: You can pick a circle as large you like in the integral formula.)
