CT Periodic Signal : $ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $
$ x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+\frac {\pi}{4})}+e^{-j(2\omega_0 t+\frac {\pi}{4})}) $
$ x(t) = 1+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j(2\omega_0 t+\frac {\pi}{4})}+\frac {1}{2j}e^{-j(2\omega_0 t+\frac {\pi}{4})} $
$ x(t) = 1e^{0j\omega_0 t}+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j\frac {\pi}{4}}e^{2\omega_0 t}+\frac{1}{2}e^{-j\frac {\pi}{4}}e^{2\omega_0 t} $
Hence we get,
$ a_0 = 1 $
$ a_1 = \frac{1}{2j}, $
$ a_{-1} = -\frac{1}{2j}, $
$ a_2 = -\frac{1}{2}e^{j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1+j), $
$ a_{-2} = -\frac{1}{2}e^{-j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1-j), $
We can write the function in the following illiterations:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $ where
$ a_3 = a_{-3} = \frac{1}{2}\, $
$ a_{4} = \frac{1}{2j} = -a_{-4}\, $
$ a_k = 0 , k \neq 3,-3,4,-4\, $
