FS of 1 + sinω₀t + 2cosω₀t + cos(2ω₀t + π/4)

Write each term using Euler's formula, with fundamental frequency $\omega_0$:

$$\begin{aligned} x(t) &= 1 + \frac{1}{2j}\left[e^{j\omega_0 t} - e^{-j\omega_0 t}\right] + \left[e^{j\omega_0 t} + e^{-j\omega_0 t}\right] + \frac{1}{2}\left[e^{j(2\omega_0 t + \pi/4)} + e^{-j(2\omega_0 t + \pi/4)}\right] \\ &= 1 + \left(1 + \frac{1}{2j}\right)e^{j\omega_0 t} + \left(1 - \frac{1}{2j}\right)e^{-j\omega_0 t} + \frac{1}{2}e^{j\pi/4}e^{j2\omega_0 t} + \frac{1}{2}e^{-j\pi/4}e^{-j2\omega_0 t} \end{aligned}$$

Comparing with $x(t) = \sum_k a_k e^{jk\omega_0 t}$ and using $\frac{1}{2j} = -\frac{j}{2}$:

$$a_0 = 1, \quad a_1 = 1 - \frac{j}{2}, \quad a_{-1} = 1 + \frac{j}{2}, \quad a_2 = \frac{1}{2}e^{j\pi/4} = \frac{\sqrt2}{4}(1 + j), \quad a_{-2} = \frac{1}{2}e^{-j\pi/4}$$

$a_k = 0$ for $|k| > 2$.

Magnitude and phase

$$|a_{\pm1}| = \sqrt{1^2 + 0.5^2} = \frac{\sqrt5}{2} = 1.118, \qquad \angle a_{\pm1} = \mp\tan^{-1}(0.5) = \mp 26.57^\circ$$

 |a_k|
 1.118     |     |
     1     |  |  |
   0.5  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

 angle a_k (deg)
    45              |
 26.57     |        |
       -+--+--o--+--+--> k
-26.57  |        |
   -45  |
       -2 -1  0  1  2

As expected for a real signal, $a_{-k} = a_k^*$: the magnitude is even and the phase is odd.