Let $\mathbb{H}$ be the upper half-plane of the complex plane. Does there exist an analytic function $A$ on $\mathbb{H}$ satisfying the inequality $\text{Re}(A(z))\ge \text{Im} (z)^{\alpha}$ for some $\alpha > 1$?
For $\alpha = 1$ this question is trivial because $A(z) = -iz$ works, however is seems that inequality of this type can't be fulfilled for any non-linear polynomial.
The answer is no since any nonnegative (actually positive if not identically zero by the usual no extremum inside a domain for real harmonic functions) harmonic function $u(x,y)$ on the upper plane cannot grow faster than linear in $y$ as $y \to \infty$ and of course, $\Re A$ is such if it satisfies the inequality $\Re A(x,y)\ge y^{\alpha}$ so, in particular, $\Re A \le Ky$ for some $K>0$ and $y \ge 1$ say.
This is an immediate consequence of the Herglotz representation formula which gives that if $u(x,y) \ge 0$ harmonic on $y>0$ then there is a (Borel) positive measure $d\mu$ on $\mathbb R$ with $\int_{-\infty}^{\infty}\frac{d\mu(t)}{1+t^2} < \infty$ and $c \ge 0$ st: $$u(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(x-t)^2+y^2}d\mu(t)+cy$$
A proof of this can be found in the usual texts on complex analysis that treat stuff like Hardy spaces on the upper half-plane in some detail, eg Garnett's Bounded Analytic Functions, page $17$ and is not that hard to do from the analog theorem on the unit disc (which is easy by using weak star compactness of positive measures with unit mass) using the standard Mobius transform to move to the upper half plane while looking at what happens at $1$ (which gets sent to infinity) carefully.
It is very easy to see that the integral part is at most $Ky$ for large $y$ and actually not that difficult to show that the integral part is $o(y)$ for large $y$ too, so $u(x,y)=cy+o(y)$ for $y \to \infty$