I have determined the complex Fourier series of the given periodic function $f(t) = e^{-t}$ over the interval $(-3,3)$ and $f(t+6)$ to be the following,
$$\sum_{n=-\infty}^\infty \left[ \frac{(-1)^n\sinh 3}{3+n\pi i}\right] e^{\pi i nt/3}$$
However I am a little confused with the following information about how to calculate the frequency spectrum, and in particular calculating
Here, the complex Fourier coefficients are, $$ c_n(f) = \frac{(-1)^n\sinh(3)}{3 + \text{i}n\pi} \, . $$ The modulus $\chi_n$ of the $n$th harmonic --which sequence is commonly denoted by amplitude spectrum-- is ${\chi_n} = \sqrt{a_n(f)^2 + b_n(f)^2}$, where \begin{aligned} a_n(f) &= \phantom{\text{i}\, (} c_n(f) + c_{-n}(f) \phantom{)} = -2\,\Im\, c_n(f)\\ b_n(f) &= \text{i} \left(c_n(f) - c_{-n}(f)\right) = \phantom{-} 2\,\Re\, c_n(f)\, . \end{aligned} Therefore, $$ {\chi_n} = 2 \left|c_n(f)\right| = \frac{2\sinh(3)}{\sqrt{9 + n^2\pi^2}}\, . $$
I am a little confused how $2|c_n|$ was calculated and am looking for some help with this calculation as I am preparing for an exam, thanks!
There are two things behind it: Euler's formula and trigonometry.
Let us discard the dependency on $f$ for sake of simplicity. We consider the partial Fourier sum $S_N$, which we rewrite using Euler's formula \begin{aligned} S_N &= \frac{a_0}{2} + \sum_{n=1}^N a_n \cos\left(\frac{n\pi t}{3}\right) + b_n \sin\left(\frac{n\pi t}{3}\right) \\ &= \frac{a_0}{2} + \sum_{n=1}^N a_n \frac{\text{e}^{n\text{i}\pi t/3} + \text{e}^{-\text{i}n\pi t/3}}{2} + b_n \frac{\text{e}^{n\text{i}\pi t/3} - \text{e}^{-\text{i}n\pi t/3}}{2\text{i}} \\ &= \frac{a_0}{2} + \frac{1}{2} \sum_{n=1}^N (a_n - \text{i}b_n)\, \text{e}^{n\text{i}\pi t/3} + (a_n + \text{i}b_n)\, \text{e}^{-n\text{i}\pi t/3} . \end{aligned} Thus, the complex Fourier coefficients satisfy $c_n = \frac{1}{2}(a_n - \text{i}b_n)$ for $n>0$, $c_0 = \frac{1}{2} a_0$ and $c_{-n} = c_n^*$. Conversely, $a_n = c_n + c_{-n}$ and $b_n = \text{i}(c_n - c_{-n})$.
Each term of the series $S_N$ is of the form $a_n\cos x + b_n\sin x$, which we may rewrite $\chi_n\sin(x-\varphi_n)$ in terms of modulus $\chi_n$ and phase $\varphi_n$ (other conventions are possible). Trigonometric relations give $$ \chi_n\sin(x-\varphi_n) = -\underbrace{\chi_n\sin\varphi_n}_{-a_n} \cos x + \underbrace{\chi_n\cos\varphi_n}_{b_n} \sin x \, . $$ Therefore, the modulus and the phase satisfy \begin{aligned} \chi_n &= \sqrt{{a_n}^2 + {b_n}^2} &&= 2 \left|c_n\right| ,\\ \tan\varphi_n &= \frac{-b_n}{a_n} &&= \frac{\Im\, c_n}{\Re\, c_n}. \end{aligned}