If $h(t)$ is the impulse response of a filter matched to a signal $s(t)$, I read that $h(t) = ks(t_o - t)$.
But what if the signal is complex? I went through the derivation of the matched filter and saw that just like how the inner product of two vectors is maximum if the vectors are "parallel". I think $H(\omega)$ has to be proportional to $S^*(\omega)e^{-\omega t_0}$ for the SNR to be maximized. This is a generalization of the Cauchy–Schwarz inequality that I don't fully understand.
So from this we can say, $H(\omega) = S^*(\omega)e^{-\omega t_0}$
To get RHS,
frequency shifting, $$s(t-t_o) \iff e^{-j\omega t_0}S(\omega)$$
taking conjugate on both sides, $$s^*(t-t_o) \iff e^{j\omega t_0}S^*(-\omega)$$
scaling with $-1$, $$s^*(t_o-t) \iff e^{-j\omega t_0}S^*(\omega)$$
So can I conclude $h(t) = s^*(t_o-t)$??
To apply this general formula to a specific example, consider this question from the 2013 exam of ISRO,
From what I learned, can I say that the answer is option D?
The impulse response for the given complex signal $s(t)$ is complex conjugate time reversal of $s(t)$, i.e. the real part of $s(t)$ simply gets time reversed (flips across y axis) so A B or D, while the imaginary part is flipped across both the y and x axis. C or D.
$$h(t) = s^*(t_o-t)$$
$$h(t) = (x(t_o-t)+iy(t_o-t))^*$$
$$h(t) = x(t_o-t)-iy(t_o-t)$$
So my question is, what is the impulse response of a matched filter if the input is complex and why? Like please point out any mistakes in my logic, I especially would love to gain some intuition on why $H(\omega)= kS^*(\omega)e^{-\omega t_0}$