CT signal:
$ x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\, $
$ x(t) = 4 * \frac{e^{j5\pi t} - e^{-j5\pi t}}{2} - (2+j)*\frac{e^{j3\pi t} + e^{-j3\pi t}}{2}\, $
$ x(t) = 2e^{j5\pi t} - 2e^{-j5\pi t} - \frac{2+j}{2}e^{j3\pi t} - \frac{2+j}{2}e^{j3\pi t}\, $
$ x(t) = 2e^{5*j\pi t} - 2e^{-5*j\pi t} - \frac{2+j}{2}e^{3*j\pi t} - \frac{2+j}{2}e^{-3*j\pi t}\, $
$ \omega_0\, $ ends up being $ \pi\, $ for this signal
So because of that, in my last step, i was able to determine what value of K's i had. (k = 3,-3,5,-5)
So then you just take the coefficients of those terms to get the $ a_k\, $
Therefore:
$ a_3 = \frac{-2-j}{2}\, $
$ a_{-3} = \frac{-2-j}{2}\, $
$ a_5 = 2\, $
$ a_{-5} = -2\, $
For every other k:
$ a_k\, = 0 $
