Analyzing Spectrum of a signal? my_signal = 8192*cos(2*pi*6*(0:7)/8) + 1i* 8192*sin(2*pi*6*(0:7)/8);

my_signal_fft = fft(my_signal);

plot(abs(my_signal_fft));


This is matlab code. What do you notice about the spectrum? Is it symmetrical? Obviously not, because I think it has an imaginary part. Would you enlighten me for the analysis of this signal's spectrum and why having an imaginary part doesn't make it symmetrical? How is the other half gone?
Your signal has the form

$$x(n)=A[\cos(2\pi fn)+i\sin(2\pi fn)]=Ae^{i2\pi fn}$$

This is a complex exponential with normalized frequency $f$ (normalized by the sampling frequency). The bins of the signal's FFT of length $N$ correspond to normalized frequencies

$$f_i=\frac{i-1}{N},\quad i=1,2,\ldots, N$$

In your case $N=8$ and $f=6/8=f_7$ (using the Matlab indexation, which starts with index $1$). So the frequency of your signal $x(n)$ lies exactly on the FFT grid. This is why there is one peak at index $7$ and all other values are $0$ (apart from numerical inaccuracies).

If $x(n)$ were real-valued, its FFT would satisfy the following symmetry condition:

$$X_k=X^*_{N-k}$$

However, for complex-valued signals, such as yours, there is in general no symmetry.