Let $f:\mathbb{C}\to \mathbb{C}$ be an entire function. Then which of the following is/are true?
$(1) \lim_{z\to \infty}\frac{f(z)}z=0$ implies $f$ is constant
$(2) \lim_{z\to \infty}\frac{Re(f(z))}z=0$ implies $f$ is constant
For $(1)$ , I think it is true and I could think of two ways of showing that.
Proof$(a)$:- Let $f(z)=a_0+a_1z+a_2z^2+...$ be the power sereis expansion of $f(z)$ .
Then for $z\neq 0$,
$\frac{f(z)}z=\frac{a_0}z+a_1+a_2z+..$
Since the $\lim_{z\to \infty}\frac{f(z)}z$ exist finitely and equals zero we must have $a_1=a_2=a_3=...=0$
Hence $f(z)=a_0$ , a constant.
Proof$(b)$ Assuming $f(z)$ is non constant entire, $z=\infty$ is either essential singularity i.e $\lim_{z\to \infty}f(z)$ does not exist (either finitely or infinitely). or $z=\infty$ is a pole singularity of some order .
If $z=\infty$ is a pole singularity, then
$0=\lim_{z\to \infty}\frac{f(z)}z[\equiv \frac{\infty}{\infty}]$
$=\lim_{z\to \infty}f'(z)$ (after L-Hospital's rule)
Since $f'(z)$ is entire and $z=\infty$ is removable singularity of $f'(z)$, so $f'(z)=0 \forall z$ and thus giving $f(z)$ is constant.
I am hoping the case for $z=\infty$ being essential singularity in proof$(b)$ to be completed but failed to do anything about it.
For the part $(2)$, I tried showing $Re(f(z)) $ is bounded and from the limit , I just have for an arbitary $\epsilon \gt 0$
$|Ref(z))|\lt \epsilon |z| ,\quad \forall z$ with $|z|\ge M$ for some $M$ .
I have no idea after this.
Do you have any suggestions for me? Thanks for your time.