If a function analytic in a strip, then its Fourier transform is bounded?

Question

Let the Fourier transform $\hat{f}(x) = \int_{\mathbb{R}}f(t)e^{ixt}dt$ exist and satisfy the relation $|\hat{f}(x)| \le C e^{-d|x|}$ for all $x \in \mathbb{R}$.Then $f$ is analytic in the strip $\left\{ z \in \mathbb{C}: |\text{Im}\ z| < d\right\}$. I can understand that if $\hat{f}$ is bounded by such exponential function, then $ f(z) = \frac{1}{2 \pi}\int_{\mathbb{R}}\hat{f}(x)e^{-ixz}dx = \frac{1}{2 \pi}\int_{\mathbb{R}}\hat{f}(x)e^{-ix(u+iv)}dx = \frac{1}{2 \pi}\int_{\mathbb{R}}\hat{f}(x)e^{-ixu}e^{xv}dx$, which is integrable with the above condition. But how to show that $f(z)$ is analytic in the strip? Is it to prove that $f$ satisfies Cauchy-Riemann equation?