Give me a hint on Stein complex analysis Chapter 4 Exercise 1 I need some hint for exercise 1 (b)

Suppose $f$ is continuous and of moderate decrease, and $\hat{f}(\xi)= 0$ for all $\xi \in \mathbb{R}$. Show that $f =0$ by completing the following outline:
(a) For each fixed real number $t$ consider the two functions
$A(z) = \int_{-\infty}^{t}f(x)e^{-2\pi i z(x-t)}dx$, $B(z) = -\int^{\infty}_{t}f(x)e^{-2\pi i z(x-t)}dx$
Show that $A(\xi) =B(\xi)$ for all $\xi \in \mathbb{R}$
(b) Prove that the function F equal to A in the closed upper half-plane, and B in the lower half-plane, is entire and bounded, thus constant. In fact, show that $F=0$.

I think I can use the theorem 3.5 in the stein's complex analysis book. The theorem states as following;

Suppose $f$ and $\hat{f}$ have moderate decrease. Then $\hat{f}(\xi) = 0$ for all $\xi < 0 $ if and only if $f$ can be extended to a continuous and bounded function in the closed upper half plane with $f$ holomorphic in the interior.

Therefore I know $f$ is holomorphic in the closed upper half plane. However, I am not sure that this fact ensure that $A(z)$ is holomorphic. Could you give me a little hint on it? If I sure it, then the rest job is to prove $F$ is holomorphic on the real line.
 A: When doing this problem, I really wanted the condition that $\hat{f}(\xi) = 0\; \forall\; x \in \mathbb{R}$. Indeed, by pulling out the textbook, this is the condition given for the problem. 
First, we get agreement of $A$ and $B$ on the real axis since $\hat{f}(\xi) = 0\; \forall \xi \in \mathbb{R}$. We then consider the extension $F$, which is bounded (since $f$ is of moderate decrease) and entire except possibly on the real-axis. Also note that $F$ is continuous everywhere since $A$ and $B$ agree on the real-axis.  Then we have that the form $f(z)\; dz$ is closed and, by Morera's Theorem, $F$ is entire. So, $F$ is both bounded and entire, whence, by Liouville's Theorem, $F$ is constant. But $|F(yi)| = |A(yi)| < \varepsilon$ for $y$ sufficiently large and so $F \equiv 0$. It then follows that $A(0) = 0$ and so by FTC, $f \equiv 0$. 
I remark that I initially thought to borrow your idea regarding Theorem 3.5, but we don't know that $\hat{f}$ is of moderate decrease. So this method seems to be a dead end.
A: 
Could you explain the reason why you say that F is entire except real axis?

Since Christopher K has not answered your comment yet, let me do it here.
First focus on the upper half-plane, where $F$ is equal to $A$. Let $A_n = \int_{-n}^t f(x) e^{2\pi i z(x-t)} dx$. $A_n(z)$ is holomorphic by Theorem 5.4 in Chapter 2. Furthermore, we see that ${A_n}$ converges uniformly to $A$ as $n$ tends to infinity. Therefore $A$ is holomorphic in the upper half-plane by Theorem 5.2 in Chapter 2.
The approach above is essentially the same as in the proof of Theorem 3.1 in the textbook.
