# 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 $z\in\Omega$.

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Let $\delta$ be a positive real number and $\tilde{f}(z)=f(z+\delta e^{i\alpha })$. It is easy to see that
$\tilde{f}(z)\in H(\Pi)$, $|\tilde{f}(z)|<1$ and
$$\lim_{r\rightarrow\infty}\frac{\log|\tilde{f}(re^{i\alpha})|}{r}=-\infty.$$
Of course $\tilde{f}(z) \in C(\bar\Pi)$.
Therefore Eclipse Sun’s argument may apply to $\tilde{g}_n(z)=\tilde{f}(z)e^{nz}$ and we have $\tilde{f}=0$, which implies $f(z)=0$ for $\operatorname{Re}\, z>\delta$. Since we can take $\delta>0$ arbitrarily (or by the coincidence theorem) we see $f=0$ for all $z\in \Pi$.

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.