Assume $f$ is holomorphic in $\{z\,|\,|z|>1\}$ and $\lim_{z\to \infty} \frac{\operatorname{Re}(f(z))}{z}=0$ Show that $\lim_{z\to \infty}f(z)$ exists.
I tried to use $h(z):=f(1/z)$ and hope to apply the removable singularity theorem at origin, but I don't know how to deal with the function $z\operatorname{Re}\bigl(h(z)\bigr)$.
We expand $f$ as Laurent series in $\{|z|>1\}$, so that $f(z)=\sum_{n=-\infty}^\infty a_nz^n$. Let $g(z)=\sum_{n=1}^\infty a_nz^{n}$. Then $g$ is an entire function such that $\lim_{z\to\infty}\frac{\mathrm{Re}g(z)}{z}=\lim_{z\to\infty}\frac{\mathrm{Re}f(z)}{z}=0$.
By this answer and question, we know that $g(z)=a_1 z$. If we plug in this expression again in the above limit, we find that $a_1=0$, and thus $g=0$. But then it follows that $f(z)=\sum_{n=-\infty}^0 a_n z^n$. The desired limit is $\lim_{z\to\infty}f(z)=a_0$.
