$A(r)=\max\limits_{|z|=r} \operatorname{Re} f(z)$, show that $\lim\limits_{ r \to \infty}\frac{\log{A(r)}}{\log{r}}=\infty$ $f$ is holomorphic in $\mathbb{C} \setminus \{ 0 \}$, $0, \infty$ are the essential singularity of $f$, let $A(r)=\max\limits_{|z|=r} \rm{Re} ~ f(z) \ (0<r<\infty)$
Please show that 
$$ \lim_{ r \to \infty} \frac{\log{A(r)}}{\log{r}}=\infty$$
$$ \lim_{ r \to \infty} \frac{\log{A(r)}}{\log{\frac{1}{r}}}=\infty$$
I use proof by contradiction. If not, there  are $\{ r_k \}, r_k \to \infty $ and $N$,satisfy $\rm{Re}~f(z) \le r_k^N, \ |z|=r_k$, but I don't know how to continue.
Thank you!
 A: I have a proof and I post it here.
We just prove the first limit here. Suppose that the Laurent series representation of $f$ is
\begin{equation}
\sum_n a_n z^n.
\end{equation}
Then,
\begin{equation}
f(re^{i\theta})=\sum_n a_n r^n e^{in\theta},
\end{equation}
giving that
\begin{equation}
\overline{f(re^{i\theta})}=\sum_n \overline{a_n}r^ne^{-in\theta}=\sum_n \overline{a_{-n}}r^{-n}e^{in\theta}.
\end{equation}
It follows that
\begin{equation}
\operatorname{Re} f(re^{i\theta})=\sum_n \frac{a_nr^n+\overline{a_{-n}}r^{-n}}{2} e^{in\theta}.
\end{equation}
Hence
\begin{equation}
\frac{a_nr^n+\overline{a_{-n}}r^{-n}}{2}=\frac{1}{2\pi}\int_0^{2\pi} \operatorname{Re} f(re^{i\theta}) e^{-in\theta} \,\mathrm{d}\theta.
\end{equation}
Suppose that there are $r_k \to \infty$ such that
\begin{equation}
\operatorname{Re} f(r_k e^{i\theta}) \le r_k^M.
\end{equation}
Then,
\begin{align}
\left\lvert \frac{a_nr_k^n+\overline{a_{-n}}r_k^{-n}}{2} \right\rvert&=\left\lvert \frac{1}{2\pi}\int_0^{2\pi} (r_k^M-\operatorname{Re} f(r_ke^{i\theta})) e^{-in\theta} \,\mathrm{d}\theta \right\rvert\\
&\le \frac{1}{2\pi}\int_0^{2\pi} (r_k^M-\operatorname{Re} f(r_ke^{i\theta})) \,\mathrm{d}\theta\\
&=r_k^M-a_0.
\end{align}
It follows that
\begin{equation}
\left\lvert \frac{a_n+\overline{a_{-n}}r_k^{-2n}}{2} \right\rvert \le r_k^{M-n}-a_0r_k^{-n}.
\end{equation}
When $n>M$, let $k \to \infty$, then we acquire that
\begin{equation}
a_n=0,
\end{equation}
which contradicts the fact that $\infty$ is an essential singularity.
