Suppose that $f$ is entire. By writing $f$ as a Taylor Series, prove that if $\lvert f(z)\rvert<m\lvert z\rvert^\alpha$ for $m>0$ and $0<\alpha<1$ then $f$ must be identically zero in $\mathbb{C}$
I'm certain this has something to do with Cauchy's integral formula for derivatives and Liouville's theorem, but I'm not sure how to apply them. If $f$ is entire, and bounded, then it must be constant, but what else must be true for it to be identically zero?
On
Write $f$ as a Taylor series: $$ f(z) = \sum_{n=0}^\infty a_n z^n \, , \quad a_n = \frac{f^{(n)}(0)}{n!} $$ The estimate $\lvert f(z)\rvert<m\lvert z\rvert^\alpha$ implies that $a_0 = f(0) = 0$. For $n\ge 1$ use Cauchy's integral formula for derivatives to estimate $$ | a_n | \le \frac{1}{r^n} \sup_{|z|=r} |f(z)| \le m r^{\alpha - n} $$ for every $r > 0$. Now take the limit $r \to \infty$ to conclude that all $a_n$ are zero.
It follows from your inequality that $f(0)=0$. Now, let$$g(z)=\begin{cases}\frac{f(z)}z&\text{ if }z\neq0\\f'(0)&\text{ otherwise.}\end{cases}$$Then $g$ is entire too and $(\forall z\in\mathbb{C}):\bigl\lvert g(z)\bigr\rvert\leqslant m\lvert z\rvert^{\alpha-1}$. Therefore, since $\lim_{z\to\infty}m\lvert z\rvert^{\alpha-1}=0$, $g$ is bounded. So, by Liouville's theorem, it is constant. That is, there is a number $K$ such that $(\forall z\in\mathbb{C}):f(z)=Kz$. But this is compatible with your inequlity only in the case in which $K=0$.