DTFS of 1 + sin + 3cos + cos(4πn/N + π/2)

Fundamental frequency $\omega_0 = 2\pi/N$. Expand each term with Euler's formula and read off the coefficients of $e^{jk\omega_0 n}$:

$$\begin{aligned} x[n] &= 1 + \frac{1}{2j}\left[e^{j\omega_0 n} - e^{-j\omega_0 n}\right] + \frac{3}{2}\left[e^{j\omega_0 n} + e^{-j\omega_0 n}\right] \\ &\quad + \frac{1}{2}\left[e^{j(2\omega_0 n + \pi/2)} + e^{-j(2\omega_0 n + \pi/2)}\right] \\ &= 1 + \left(\frac{3}{2} + \frac{1}{2j}\right)e^{j\omega_0 n} + \left(\frac{3}{2} - \frac{1}{2j}\right)e^{-j\omega_0 n} \\ &\quad + \frac{1}{2}e^{j\pi/2}e^{j2\omega_0 n} + \frac{1}{2}e^{-j\pi/2}e^{-j2\omega_0 n} \end{aligned}$$

Fourier series coefficients (one period, $k = -2$ to $2$)

Using $\frac{1}{2j} = -\frac{j}{2}$:

Working: $|a_1| = \sqrt{1.5^2 + 0.5^2} = \sqrt{2.5} = 1.581$, $\angle a_1 = -\tan^{-1}(0.5/1.5) = -18.43^\circ$ ($-0.322$ rad).

The coefficients repeat with period $N$: $a_{k+N} = a_k$ (e.g. $a_{N-1} = a_{-1}$, $a_{N-2} = a_{-2}$). Here $N \ge 5$ is assumed so that the harmonics do not overlap.

Plots (one period, $k = -2 \dots 2$, repeating every $N$)

 |a_k|  (repeats every N)
 1.581     |     |
     1     |  |  |
   0.5  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

 angle a_k (deg)
    90              |
 18.43     |        |
       -+--+--o--+--+--> k
-18.43  |        |
   -90  |
       -2 -1  0  1  2

Magnitude is even in $k$ and phase is odd in $k$, as expected for a real signal ($a_{-k} = a_k^*$).