Q:

Can’t understand the complex notation for trigonometric polynomials

A reference from “Introduction to calculus and analysis I”:
A compound vibration of the type
$$
S_n(x)=\frac{a_0}{2}+\sum_{v=1}^{n}(a_v\cos{vx}+b_v\sin{vx})
$$
(For brevity we have taken $\omega=1$) can be reduced to complex form by substituting 
$$
\cos{vx} = \frac{1}{2}(e^{ivx}+e^{-ivx}),{\quad}\sin{vx}=-\frac{1}{2}i (e^{ivx}-e^{-ivx}).
$$
This expression then assumes the simpler form 
$$
S_n(x)=\sum_{v=-n}^{n}{\alpha}_{v} e^{ivx},
$$
where the complex numbers ${\alpha}_{v}$ are related to the real numbers $a_0$, $a_v$, $b_v$ by the equations 
$$
\alpha_v = \frac{1}{2}(a_v-ib_v), \\
\alpha_{-v} = \frac{1}{2}(a_v+ib_v),  \\
\alpha_{0}=\frac{1}{2}a_0.
$$
For $v=1,2,\dots,n$, and solving these relations for the $a_v$ and $b_v$, we find that
$$
a_v=\alpha_v+\alpha_{-v},\\
b_v=i(\alpha_v-\alpha_{-v}).
$$
(This case $v=0$ is included.)
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Please explain it to me, thanks.

A:

The $2$ complex forms for $\cos vx$ and $\sin vx$ you are given come from Euler's formula which states
$$e^{ix} = \cos x + i\sin x \tag{1}\label{eq1}$$
Using this gives that
$$\begin{equation}\begin{aligned}
\frac{1}{2}(e^{ivx}+e^{-ivx}) & = \frac{1}{2}(\cos(vx) + i\sin(vx) + \cos(-vx) + i\sin(-vx)) \\
& = \frac{1}{2}(\cos(vx) + i\sin(vx) + \cos(vx) + -i\sin(vx)) \\
& = \frac{1}{2}(2\cos(vx)) \\
& = \cos(vx)
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
Similarly, you get
$$\sin(vx) = -\frac{1}{2}i (e^{ivx}-e^{-ivx}) \tag{3}\label{eq3}$$ 
Substituting those $2$ forms into the summation, and collecting the $e^{ivx}$ and $e^{-ivx}$ terms, gives
$$\begin{equation}\begin{aligned}
S_n(x) & = \frac{a_0}{2}+\sum_{v=1}^{n} \left(a_v\left(\frac{1}{2}(e^{ivx}+e^{-ivx})\right) + b_v\left(-\frac{1}{2}i (e^{ivx}-e^{-ivx})\right)\right) \\
& = \frac{a_0}{2} + \sum_{v=1}^{n}\left(\frac{1}{2}(a_v - ib_v)e^{ivx} + \frac{1}{2}(a_v + ib_v)e^{-ivx} \right) \\
& = \frac{a_0}{2} + \sum_{v=1}^{n}\left(\frac{1}{2}(a_v - ib_v)e^{ivx}\right) + \sum_{v=-n}^{-1}\left(\frac{1}{2}(a_{-v} + ib_{-v})e^{ivx}\right) \\
& = \sum_{v=-n}^{n}{\alpha}_{v} e^{ivx}
\end{aligned}\end{equation}\tag{4}\label{eq4}$$
You have the initial constant term for $v = 0$ of $\alpha_0 = \frac{1}{2}a_0$. Also, wth the $e^{ivx}$ terms, you get $\frac{1}{2}(a_v-ib_v)$, which is the $\alpha_v$. With the $e^{-ivx}$ terms, you get the $\frac{1}{2}(a_v+ib_v)$, which is the $\alpha_{-v}$ (with the "$-$" for the fact the index is negative).
At the end, you can solve for the terms by adding the $\alpha_v = \frac{1}{2}(a_v-ib_v)$ and $\alpha_{-v} = \frac{1}{2}(a_v+ib_v)$ expressions to get $a_v=\alpha_v+\alpha_{-v}$. Also, subtracting the second from the first you get $ib_v = (-\alpha_v+\alpha_{-v}) \implies -b_v = i(-\alpha_v+\alpha_{-v}) \implies b_v = i(\alpha_v-\alpha_{-v})$, where I used $i^2 = -1$.
Finally, the $v = 0$ case is included because the $2$ equations give $a_0 = \alpha_0 + \alpha_0 = \frac{1}{2}a_0 + \frac{1}{2}a_0 = a_0$, and $b_0 = i(\alpha_0 - \alpha_0) = 0$, which is correct as there's no $b_0$ term being used.

