Cfrgtkky

Let
$$Theta(z|tau) = sum_{n = -infty}^{infty}e^{pi in^{2}tau}e^{2pi inz}$$
with $tau = s + it$ where $t > 0$. Show that $Theta$ is of order $2$ as function of $z$.

[Hint: $-n^{2}t + 2n|z| leq -n^{2}t/2$ when $t>0$ and $n geq 4|z|/t$.]

I dont know if its necessary to describe all my calculations. Initially, I will write where my problem is. I got:

$$|Theta(z|tau)| leq K + underbrace{2sum_{n=1}^{alpha}e^{-pi n^{2}t + 2pi n|z|}}_{S},$$

where $displaystyle alpha = leftlfloor frac{4|z|}{t} rightrfloor$.

Question 1:

The sum $S$ is finite, there is at most $alpha$ terms. So, I'm trying to bound each term of the sum. I know that if cannot to eliminate $|z|$. I thought about to use $nleq 4|z|/t$ for conclude that
$$|Theta(z|tau)| leq K_{1} + K_{2}e^{K_{3}|z|^{2}},$$
but I couldnt prove it.

Question 2:

Assuming I have
$$|Theta(z|tau)| leq K_{1} + K_{2}e^{K_{3}|z|^{2}}.$$

How can I use this to get
$$|Theta(z|tau)| leq Ae^{B|z|^{2}}?$$

EDIT. I can prove that
$$|Theta(z|tau)| leq Ae^{B|z|^{2}}.$$
Just note that $e^{-pi n^{2}t + 2pi n|z|} leq e^{2pi n|z|}$ and use the hypothesis about $n$. But with this, I show that $Theta$ is of order at most $2$. How can I show that the order is exactly $2$?

Definition. A entire function $f$ such that
$$|f(z)| leq Ae^{B|z|^{rho}}tag{$ast$}$$
for some $rho,A,B>0$ and for any $z in mathbb{C}$ is of order $leq rho$. The order of $f$ is $rho_{f} = inf rho$ for all $rho$ that satisfies $(ast)$.