Reading Tao's proof of Hardy's Uncertainty principle, I found the claim that
$$e^{i\epsilon e^{i\epsilon}z^{2+\epsilon}}e^{i\delta z^2}F(z) \to 0$$ as $z\to\infty$ inside the sector $\{re^{i\alpha}:r>0, 0\leq\alpha\leq\theta\}$, where $\epsilon,\delta>0$ and $\theta\in(0,\pi/2)$ is close to $\pi/2$.
The function $F(z)=e^{\pi z^2}\hat{f}(z)$, and $f$ is a function such that $f$ and $\hat{f}$ are bounded by Gaussians $e^{-\pi x^2}$ on the real line.
Why does the above expression converge to 0?
EDITED
I previously claimed that
$$ |e^{i\epsilon e^{i\epsilon}z^{2+\epsilon}}e^{i\delta z^2}F(z)| \to 0$$
as $r\to\infty$ because $|e^{i\epsilon e^{i\epsilon}z^{2+\epsilon}}|\to 0$ and $|e^{i\delta z^2}F(z)|$ remains bounded in $\Gamma_\theta^\circ$. This is circular, since at this point we are trying to prove precisely the last claim in italics; it has only been shown that $|e^{i\delta z^2}F(z)|\leq1$ on the boundary of $\Gamma_\theta^\circ$. The following is a revised and corrected argument.
Points in the interior of $\Gamma_{\theta}$ are of the form $z = re^{i\alpha}$ with $0<\alpha<\theta$. Hence, $$ z^{2+\epsilon} = r^{2+\epsilon}e^{i(2+\epsilon)\alpha}$$ and we can compute $$ w := e^{i\epsilon}z^{2+\epsilon} = r^{2+\epsilon}e^{i(2\alpha+\alpha\epsilon+\epsilon)}.$$ Geometrically, we obtain a point $w$ further out from $O$ (at $r^{2+\epsilon}$) and still in the upper half-plane, provided $0<2\alpha+\alpha\epsilon+\epsilon<\pi$, i.e. when $\epsilon <\frac{\pi-2\alpha}{\alpha+1}$. Thus, it suffices to take $\epsilon < \pi-2\theta$, which is positive and independent of $\alpha$.
Next, note that the expression
$$e^{i\epsilon e^{i\epsilon}z^{2+\epsilon}}e^{i\delta z^2}F(z)$$
is continuous in $z$ in the interior of $\Gamma_\theta$. Thus, we may keep $\alpha$ fixed and take $r\to+\infty$ to compute the limit as $z\to\infty$ in the interior of $\Gamma_\theta$.
Since we are given that $|F(z)|\leq e^{\pi\operatorname{Re}(z)^2}$, and $\operatorname{Re}(z)=r\cos(\alpha)$,
$$|e^{i\delta z^2}F(z)| \leq e^{-2\delta r^2+\pi r^2\cos^2(\alpha)} \leq e^{Br^2}$$
where $B=\pi\cos^2(\alpha)\geq 0$. Hence
$$|e^{i\epsilon e^{i\epsilon}z^{2+\epsilon}}e^{i\delta z^2}F(z)| \leq e^{-Ar^{2+\epsilon}+Br^2} = e^{r^2(A-Br^\epsilon)}$$
where $A=\epsilon\sin^{2+\epsilon}(2\alpha+\alpha\epsilon+\epsilon)>0$, has a vanishing modulus as $r\to+\infty$. Note that $A$ and $B$ are positive and only depend on the fixed quantities $\epsilon>0$, $0<\alpha<\theta$.