Let $f$ be a function holomorphic at $1<|z|<\infty$ and $z^{-1}\operatorname{Re}f(z)\to 0$ as $z\to \infty$. Prove that $\lim_{z\to \infty} f(z)\in \mathbb{C}$ exists.
I want to prove this by first showing function holomorphic on $\mathbb{C}$ with the same condition reduces to a constant but don't know how to proceed. Thanks for any help.
Let $f(z)=\sum_{n=-\infty}^\infty a_nz^n$ be its Laurent expansion in $1<|z|<\infty.$ Then \begin{align} &u(Re^{i\theta })=\operatorname{Re}f(Re^{i\theta })\\ &=\operatorname{Re}a_0+\sum_{n=1}^\infty \left[\left(\operatorname{Re}a_nR^n+\operatorname{Re}a_{-n}R^{-n}\right)\cos n\theta -\left(\operatorname{Im}a_nR^n-\operatorname{Im}a_{-n}R^{-n}\right)\sin n\theta \right] \end{align}
Therefore for $n\ge 1$, we have \begin{align} &\operatorname{Re}a_nR^n+\operatorname{Re}a_{-n}R^{-n} =\frac{1}{\pi}\int_0^{2\pi}u(Re^{i\theta })\cos n\theta \,d\theta, \\ &\operatorname{Im}a_nR^n-\operatorname{Im}a_{-n}R^{-n}=-\frac{1}{\pi}\int_0^{2\pi}u(re^{i\theta })\sin n\theta \, d\theta. \end{align} Dividing by $R^n$ we have \begin{align} &\operatorname{Re}a_n+\operatorname{Re}a_{-n}R^{-2n} =\frac{1}{\pi}\int_0^{2\pi}\frac{u(Re^{i\theta })}{R^n}\cos n\theta \,d\theta, \tag{1} \\ &\operatorname{Im}a_n-\operatorname{Im}a_{-n}R^{-2n}=-\frac{1}{\pi}\int_0^{2\pi}\frac{u(re^{i\theta })}{R^n}\sin n\theta \, d\theta.\tag{2} \end{align} By the condition $z^{-1} \operatorname{Re}f(z)\to 0$ as $z\to \infty$, we have $$\frac{1}{R}u(Re^{i\theta })\to 0 \:\;(R\to \infty). $$ Tending $R\to \infty$ in $(1),(2)$ we have $$ \operatorname{Re}a_n =0,\quad \operatorname{Im}a_n=0, $$ that is, $a_n=0$ for $n\ge 1.$ Thus $$ f(z)=a_0+\sum_{n=1}^\infty \frac{a_{-n}}{z^n}.$$ We see that $\lim_{z\to \infty}f(z)$ exists.