If $$x_{in}(t) = \sin(2 * \pi * 4000 * t) + 0.75 * \sin \left(2 * \pi * 5000 * t + \frac{\pi}{4}\right)$$ is sampled with $$F_s = 16000 Hz$$ calculate $$X(0)$$ if $$X(m) = \sum_{n = 0}^{N - 1} x(n) e^{-j2 \pi nm/N}$$ When $$N = 8$$, where $$x(n) = x_{in}(nt_s)$$

A

0.0 - j 4.0

B

0.0 - j 0.0

C

1.414 + j 1.414

D

0.0 + j 4.0