Complex analysis, showing a function is constant Let $\Omega$ be the right half plane excluding the imgainary axis and $f\in H(\Omega)$ such that $|f(z)|<1$ for all $z\in\Omega$. If there exists $\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})$ such that $$\lim_{r\rightarrow\infty}\frac{\log|f(re^{i\alpha})|}{r}=-\infty$$ prove that $f=0$.
The hint is define $g_n(z)=f(z)e^{nz}$, then by previous exericise $|g_n|<1$ for all $z\in\Omega$.

I can only give a partial proof.  
Let $g_n(z)=f(z)e^{nz}$, $n=1,2,\ldots$, and $g_n(z)\in H(\Pi)$.  
Assume $g_n(z)\in C(\bar\Pi)$. 
Let $K_1=\{re^{i\theta}\mid\theta\in(\alpha,\pi/2)\}$, $K_2=\{re^{i\theta}\mid-\theta\in(\alpha,\pi/2)\}$ and $K_3=\Pi\setminus(K_1\cup K_2)$. By the assumption $\lim_{r\rightarrow\infty}\frac{\log|f(re^{i\alpha})|}{r}=-\infty$, $g_n(z)$ is bounded on each sector $K_i$. Since $g_n(z)$ doesn't grow too fast, by Phragmen-Lindelöf's principle for sectors, $g_n(z)$ is bounded in $\Pi$.  
By Phragmen-Lindelöf's principle again, this time for the right half plane, $|g_n(z)|\le1$. Since it is true for all $n$, we must have $f=0$.  
It remains to show that the assumption $g_n(z)\in C(\bar\Pi)$ is true, or it is redundant.
