= Solution
Write $h(z)=\sum_{n=0}^{\infty}a_nz^n$. Cauchy's coefficient formula on $|z|=R$ gives
$$
a_n=\frac1{2\pi R^n}
\int_0^{2\pi}h(Re^{i\theta})e^{-in\theta}\,d\theta.
$$
Except at the two measure-zero points where $\cos\theta=0$, the hypothesis gives
$$
|h(Re^{i\theta})|
\leq R^{-1/2}|\cos\theta|^{-1/2}.
$$
Since $|\cos\theta|^{-1/2}$ is integrable,
$$
|a_n|
\leq C R^{-n-1/2}
$$
for a constant $C$ independent of $R$. Letting $R\to\infty$ shows that every $a_n=0$. Hence $h\equiv0$, in particular $h$ is constant. This proves the <entire function under a horizontal inverse-square-root bound> result.
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