What will the impulse response of a matched filter look like if the input is complex?
The impulse response doesn't change because it is the output of the filter when the input is an impulse.

When the complex input is applied to a complex filter the output is still determined by the convolution of the input and the impulse response of the filter. The addition/integration and multiplication operations that take place inside a convolution operation have to account for complex numbers.

I think it is difficult to gain intuition for the result of a matched filter in the time domain. Instead look at it from the frequency domain.
\begin{align}
 Y(\omega) &amp;=&amp;  H(\omega)S(\omega) \\
 &amp; = &amp;  e^{-j\omega t_0}S^*(\omega) S(\omega)\\
 &amp; = &amp;  e^{-j\omega t_0} ||S(\omega)||^2
\end{align}

In the case where the envelope of $S(\omega)$ is rectangular (as in your example) then $||S(\omega)||^2$ is equal to a rectangular pulse. The output then be a time delayed version of a sinc() pulse. The wider the frequency range of $S(\omega)$ then the narrower the main pulse of the sinc() pulse.

If you are trying to gain insight into how a matched filter is derived - one insight I can provide is that you have to understand that the Signal to Noise ratio (SNR) is defined differently at the input of the filter, than the SNR that we are trying to maximize at the output. At the output we are interested in the maximum value of the output signal, while at the input the SNR is measured using the signal's energy.